iiiiiliilii 


ESSENTIALS   OF   / 

ELECTRICAL  ENGINEERING 

A    TEXT    BOOK    FOR    COLLEGES   AND 
TECHNICAL   SCHOOLS 


BY 


JOHN   FAY   WILSON,   B.  S.,  E.  E. 

Instructor  in  Electrical  Engineering  at  the  University  of  Michigan 


282  ILLUSTRATIONS 


NEW  YORK 

D.  VAN  NOSTRAND  COMPANY 

25  PARK  PLACE 
1915 


r  K 


COPYRIGHT,  1915, 

BY 
D.  VAN  NOSTRAND   COMPANY 


Stanbopc  ipress 

f,   H.  CILSON     COMPANY 
BOSTON.     U.S.A. 


PREFACE 


The  widely  prevalent  belief  that  continuous  and  alternating  cur- 
rents are  not  subject  to  the  same  general  laws,  is  entirely  errone- 
ous. The  principles  and  laws  which  relate  to  the  flow  of  continu- 
ous currents  also  govern  the  flow  of  alternating  currents. 

This  volume,  which  is  offered  as  a  text  for  the  use  of  students 
pursuing  either  electrical  or  non-electrical  engineering  courses,  is 
the  result  of  the  writer's  class-room  experience,  and  seeks  to  em- 
phasize the  fact  that  continuous  and  alternating  currents  are  gov- 
erned by  the  same  laws.  To  this  end  the  fundamental  laws  of  the 
electric  circuit  are  fully  developed  before  any  study  of  machines  is 
attempted.  With  a  thorough  knowledge  of  the  electric  circuit 
as  a  foundation,  the  student  should  have  little  trouble  in  compre- 
hending the  physical  phenomena  taking  place  in  the  more  common 
types  of  electrical  apparatus. 

The  student  is  expected  to  be  familiar  with  trigonometry,  and  a 
knowledge  of  calculus  will  be  found  advantageous  but  not  indis- 
pensable. The  mathematical  developments  of  the  formulae  for  the 
calculation  of  inductance  and  capacitance  have  been  placed  in 
appendices  at  the  back.  These  and  other  portions  of  the  text  may 
be  omitted  when,  for  lack  of  time  or  for  any  other  reason,  it  is 
necessary  to  shorten  the  course. 

The  fact  that  the  ideas  advanced  in  this  volume  have  developed 
with  the  science  of  Electrical  Engineering,  and  may  be  regarded 
as  the  common  property  of  the  science,  would  make  any  attempt 
to  give  specific  credit  burdensome  (and  often  impossible),  but  the 
writer  wishes  to  specifically  acknowledge  his  indebtedness  to  both 
standard  and  current  literature,  particularly  to  those  works  listed 
on  page  333.  Students  desiring  a  more  detailed  discussion  of 
particular  subjects  are  referred  to  this  list. 

The  writer  also  wishes  to  express  his  obligation  to  the  following 
men,  each  of  whom  read  all  or  part  of  the  manuscript,  and  offered 
valuable  suggestions  for  its  improvement:  Professor  C.  M.  Jan- 

337873 


iv  PREFACE 

sky,  of  the  University  of  Wisconsin;  Professor  H.  H.  Higbie, 
Professor  A.  H.  Lovell,  Mr.  A.  H.  Stang  and  Mr.  W.  L.  Bice,  of 
the  University  of  Michigan. 

Some  idea  of  the  practical  construction  of  electrical  machinery 
and  instruments  is  given  by  means  of  a  limited  number  of  illustra- 
tions of  actual  apparatus.  These  illustrations  are  offered  through 
the  courtesy  of  the  manufacturers. 

J.  F.  W. 

ANN  ARBOR,  MICH. 
June,  1915. 


TABLE   OF   CONTENTS 


CHAPTER  PAGES 

I.  THE  ELECTRIC  CIRCUIT 1-28 

1  II.  MAGNETISM  AND  MAGNETIC  INDUCTION 29-45 

III.  PRACTICAL  CONSTRUCTION  OF  THE  DYNAMO 46-59 

IV.  THE  CONTINUOUS-CURRENT  GENERATOR 60-78 

V.  THE  CONTINUOUS-CURRENT  MOTOR 79-92 

VI.  LOSSES,   EFFICIENCIES  AND   RATINGS  OF   CONTINUOUS-CURRENT 

DYNAMOS 93-106 

VII.  POLYPHASE  ALTERNATING  CURRENTS 107-122 

VIII.  THE  ALTERNATING-CURRENT  GENERATOR 123-148 

IX.  THE  SYNCHRONOUS  MOTOR 149-160 

X.  CURRENT-RECTIFYING  APPARATUS 161-177 

XI.  THE  TRANSFORMER 178-195 

XII.  TRANSFORMER  CONNECTIONS . 196-202 

XIII.  THE  INDUCTION  MOTOR 203-225 

XIV.  SINGLE-PHASE  COMMUTATING  MOTORS 226-234 

XV.  ELECTRIC  LAMPS 234-243 

XVI.    CURCUIT-LNTERRUPTING  APPARATUS '. 244-25! 

XVII.  METERS 252-267 

XVIII.  POWER  TRANSMISSION  AND  DISTRIBUTION 268-298 

XIX.  THE  STORAGE  BATTERY 299-304 

APPENDIX  A.  HARMONIC  QUANTITIES 305-310 

APPENDIX  B.  INDUCTANCE 311-318 

APPENDIX  C.  CAPACITANCE 319-327 

APPENDIX  D.  THE  COMPLEX  QUANTITY.    ADMITTANCE,  CONDUCTANCE 

AND  SUSCEPTANCE 328-330 

APPENDIX  E.  RESUSCITATION  FROM  ELECTRIC  SHOCK 33I-332 

REFERENCES..                                                                                   333 


NOTATION* 


*/  A  =  area. 

v  B  =  magnetic  flux  density. 

b  =  susceptance. 

v    C  =  capacitance  (electrostatic  capacity). 
r  D  =  electrostatic  flux  density. 

•^  e  —  electromotive  force,  instantaneous  alternating  electromotive  force. 
*/  Em  =  maximum  alternating  electromotive  force. 

•*  E  =  continuous  electromotive  force,  effective  alternating  electromotive  force. 
-v     /  =  frequency,  force. 

5~  =  magnetomotive  force. 
^  F  =  electrostatic  field  intensity. 
-v  g  =  conductance. 
T    E  =  magnetic  field  intensity,  magnetizing  force. 

i  =  instantaneous  current. 
•»-    Im  =  maximum  alternating  current. 

/  =  continuous  current,  effective  alternating  current. 
K  =  dielectric  constant. 
kva  =  kilovolt-amperes. 
kw  =  kilowatts. 
L  =  inductance. 
/  =  length. 

m  =  unit  magnet  pole. 
n  =  speed  (revolutions  per  second). 

N  =  number  of  armature  conductors,  number  of  turns  in  a  winding. 
P  =  power  in  continuous-current  circuit,  average  power  in  alternating-current 

circuit. 
p  =  number  of  poles  on  a  dynamo,  instantaneous  power  in  alternating-current 

circuit. 

P'  =  number  of  paths  into  which  an  armature  winding  is  divided. 
q,  Q  =  quantity  of  electricity. 
R  =  resistance. 
$1  =  reluctance. 
s  =  slip  of  an  induction  motor. 
t  F=  time. 

T  =  torque,  temperature. 
V  =  velocity,  volume. 
W  =  work. 
X  =  reactance. 
Y  =  admittance. 
Z  =  impedance. 
a  =  an  angle. 

*  Based  on  the  report  of  the  Standardization  Committee  of  the  American  Institute 
of  Electrical  Engineers  (1915). 

vii 


viii  NOTATION 

/3  =  an  angle. 

<f>  =  magnetic  flux,  an  angle. 

0  =  an  angle. 

^  =  dielectric  flux. 

p  —  permeability. 

co  =  angular  velocity  (radians  per  second). 

p  =  resistivity. 


Essentials  of  Electrical  Engineering 


CHAPTER  I 
THE  ELECTRIC   CIRCUIT 

1.  Introduction.  —  The  ultimate  nature  of  electricity  has  never 
been  discovered  but  it  is  generally  conceived  to  be  a  medium, 
without  weight  or  form,  by  means  of  which  the  energy  of  heat  or 
motion  may  be  transferred  from  one  point  to  another.     By  means 
of  this  medium  the  energy  of  motion,  as  developed  by  the  electric 
generator,  may  be  transferred  and  caused  to  reappear  at  some  dis- 
tant point  in  the  form  of  motion  (the  electric  motor),  or  as  heat  (the 
electric  lamp).     This  conception  of  electricity  is  upheld  by  the 
similarity  of  an  electric  and  a  hydraulic  system. 

Contrary  to  a  widely  prevalent  belief,  electricity  is  not  erratic  in 
its  action  but  is  governed  by  simple  and  well-defined  laws.  In  the 
following  pages  the  fundamental  laws  of  the  electric  circuit  are  de- 
veloped, and  the  physical  phenomena 
which  take  place  in  the  more  common 
types  of  electrical  apparatus  explained. 

2.  The  electric  current.  —  The  elec- 
tric circuit  (Fig.  i)  is  analogous,    in 
many  respects,  to  a  hydraulic  system, 

consisting  of  the  pump  and  pipe  connections  shown  in  Fig.  2. 

The  rotary  pump  indicated  in  Fig.  2a  produces  a  continuous  and 
, __ ^  uniform  flow  of  water  in  the  direction  indi- 
cated by  the  arrows;  the  piston  pump  indicated 
in  Fig.  2b  produces  a  flow  which  is  always  in 
the  direction  indicated  by  the  arrows  but 
which  is  not  uniform,  i.e.,  when  the  piston 


Generator 


FIG.  i.    The  Electric  Circuit. 


FIG.  2a.   Hydraulic  Anal-  speed  is  reduced  and  its  motion  reversed  at  the 
ogy    for    Continuous  end  of  the  stroke,  the  flow  of  water  slackens  or 
ceases  altogether;   the  valveless  pump  indi- 
cated in  Fig.  2c  produces  a  flow  which  is  not  constant,  either  in 
direction  or  in  value,  but  surges  first  in  one  direction  and  then  in 
the  other  through  the  system. 


ENGINEERING 


Similarly  the  construction  of  the  generator  determines  the  char- 
acteristics of  the  electric  current  flowing  in  the  circuit.     Electric 

currents  may  be  divided  into 

s —  ^T  — x  » 

three   classes:   (a)  continuous, 
(b)  pulsating,  (c)  alternating. 

(a)  Continuous  currents.  — 
When  the  current  in  a  given 
circuit  flows  continuously  in 
one  direction  and  the  rate  of 
flow  is  uniform,  it  is  said  to  be 
a  " continuous  current"  and  is 


Piston 
.-•'  Pump 


FIG.  2b. 


Hydraulic  Analogy  for  Pulsating 
Currents. 


commonly  called  a  "  direct 
current."  The  voltaic  cell 
and  certain  forms  of  the  electric  generator  cause  a  unidirectional 
current  to  flow  at  an  approximately  constant  rate  and  are,  there- 
fore, continuous-current  appa- 
ratus. 

(b)  Pulsating  currents.  — 
If  the  current  in  a  given  cir- 
cuit flows  in  one  direction 
but  at  a  momentarily  chang- 
ing rate,  it  is  a  "pulsating 
current."  Pulsating  currents 


Valveless 
Pump 


FlG.   2C. 


Hydraulic  Analogy  for  Alternating 
Currents. 


are  largely  used  in  telephone 
work  and  in  telegraphy. 

(c)  Alternating  currents.  —  When  the  current  in  a  given  circuit, 
starting  at  zero,  increases  to  maximum,  decreases  to  zero,  increases 
to  maximum  in  the  opposite  direction,  and  again  decreases  to  zero 
(the  process  being  repeated  periodically),  it  is  an  "alternating  cur- 
„  rent."  The  wave  forms  produced  by  commercial  alternators  vary 
greatly,  but  comparisons  and  mathematical  calculations,  unless 
specifically  stated  otherwise,  are  based  on  the  harmonic  or  sine  wave 
(  form.*  While  this  wave  form  is  seldom  or  never  attained,  its  as- 
sumption is  usually  sufficiently  accurate  for  practical  purposes. 
Calculations  for  the  actual  form  of  the  wave  may  become  extremely 
complicated. 

3.  Manifestations  of  the  electric  current.  —  The  presence  of  an 
electric  current  is  made  manifest  in  a  number  of  ways.     The  prin- 

*  See  Appendix  A. , 


THE   ELECTRIC   CIRCUIT  3 

cipal  effects  of  the  electric  current  with  which  the  electrical  engi- 
neer is  concerned  are:  (a)  chemical,  (b)  magnetic  and  (c)  heating. 

(a)  Chemical  effect.  —  When  water  containing  a  small  quantity 
of  acid  forms  part  of  a  circuit  carrying  a  unidirectional  current, 
bubbles  may  be  seen  to  rise  through  the  liquid.     When  the  gases 
causing  these  bubbles  are  collected,  they  are  found  to  be  oxygen  and 
hydrogen,  the  elements  which  enter  into  chemical  combination  to 
form  water.     The  same  action  takes  place  in  a  circuit  carrying 
alternating  current,  but  each  half  cycle  destroys  the  chemical  effect 
of  the  preceding  half  cycle  so  that  the  net  effect  is  zero.     Hence, 
alternating  currents  are  not  used  when  chemical  effects  are  desired. 

(b)  Magnetic  effect.  —  A  current-carrying  conductor  exhibits  all 
the  characteristics  of  a  magnet  in  that  it  attracts  pieces  of  iron, 
and  is  attracted  to  or  repelled  from  a  magnet  or  another  current- 
carrying  conductor.     Attraction  or  repulsion  takes  place  accord- 
ing to  the  polarity  of  the  magnet,  or  the  relative  directions  of  the 
currents  in  the  conductors.     This  magnetic  effect  is  intensified  if 
the  conductor  is  in  the  form  of  a  spiral.     It  is  still  further  increased 
if  the  spiral  is  wound  about  an  iron  core. 

(c)  Heating  effect.  —  When  an  electric  current  flows  in  a  wire,  heat 
is  liberated.     The  heating  may  be  so  slight  as  to  be  unnoticed  as  in 
the  ordinary  electric  bell  circuit,  or  so  great  as  to  heat  the  conductor 
to  incandescence  as  in  the  electric  glow  lamp.     Light  is  not  a  direct 
manifestation  of  the  electric  current,  but  is  due  to  the  high  tempera- 
ture to  which  the  current  heats  the  conductor. 

4.  Series  circuits.  —  A  series  circuit,  represented  in  Fig.  3a,  is 
one  in  which  the  entire  current  in  the  circuit  flows  successively 
through  each  piece  of  apparatus.  The  commercial  application  of 
the  series  circuit  is  somewhat  limited,  the  most  common  use  being 
in  connection  with  street  and  other  outdoor  lighting. 


n 

A  A          e-L 


Load  A  A          e-Load 


Co)  (b) 

FIG.  3.    Series  and  Parallel  Circuits. 


5.  Parallel  circuits.  -7-  A  parallel  or  multiple  circuit,  represented 
in  Fig.  3b,  is  one  in  which  the  current  divides  and  flows  through 
two  or  more  branches.  The  current  in  each  branch  of  a  parallel 
circuit  is  independent  of  that  in  the  other  branches.  Motors  are 
always  connected  in  parallel. 


4  ESSENTIALS   OF  ELECTRICAL  ENGINEERING 

6.  Electric  units.*  -  -  The  units  of  the  electric  system  are :  (a)  the 
ohm,  (b)  the  ampere,  (c)  the  volt,  (d)  the  coulomb,  (e)  the  joule  and 
(/)  the  watt. 

(a)  The  ohm.  —  The  opposition  to  the  flow  of  an  electric  current 
is  measured  in  ohms.     The  standard  ohm  is  represented  by  the 
opposition  offered  to  an  unvarying  electric  current  by  a  column 
of  mercury  at  a  temperature  of  melting  ice,   14.4521   grams  in 
mass,  of  a  constant  cross  section,  and  of  the  length  106.3  centi- 
meters. 

(b)  The  ampere.  —  The  rate  at  which  electricity  is  transferred  is 
measured  in  amperes,  analogous  to  miner's  inches  or  gallons  per 
second,  and  the  ampere  is  represented  by  the  unvarying  current 
which,  when  passed  through  a  standard  solution  of  nitrate   of 
silver  in  water,  deposits  silver  at  the  rate  of  0.001118  gram  per 
second. 

(c)  The  volt.  —  The  force  which  causes  or  tends  to  cause  an  elec- 
tric current  to  flow  is  termed  an  " electromotive  force,"  the  unit  of 
which,  the  volt,  is  the  electromotive  force  that,  steadily  applied 
to  a  conductor  whose  resistance  is  one  ohm,  produces  a  current  of 
one  ampere. 

(d)  The  coulomb.  —  Quantity  of  electricity  is  expressed  in  cou- 
lombs, analogous  to  gallons  or  cubic  feet,  and  one  coulomb  is 
the  quantity  of  electricity  transferred  in  one  second  by  a  current 
of  one  ampere. 

(e)  The  joule.  —  The  unit  of  energy  or  work  in  the  electric  system 
is  the  joule,  and  is  the  energy  expended  in  one  second  when  a  current 
of  one  ampere  flows  in  a  circuit  having  a  resistance  of  one  ohm. 
The  energy  equivalent  of  one  joule  has  been  experimentally  deter- 
mined to  be 

==  0.24  calorie. 
=  0.00095  B.T.U. 
=  0.737  foot-pound. 
=  10,000,000  ergs. 

(/)  The  watt. — The  watt,  which  is  the  unit  of  electrical  power,  is 
represented  by  the  expenditure  of  one  joule  in  one  second.  Since 

*  There  is  a  definite  relation  between  the  C.G.S.  units  (the  so-called  absolute  units) 
and  the  ohm,  the  ampere  and  the  volt,  but  these  units  are  to  be  regarded  as  arbitrary 
units,  similar  to  the  foot  and  the  pound,  which  have  been  adopted  by  international 
agreement,  and  legalized  by  statutory  enactment. 


THE   ELECTRIC    CIRCUIT  5 

the  energy  equivalent  of  one  joule  is  0.737  foot-pound,  and  550 
foot-pounds  per  second  equal  one  horse  power, 

h.p.  =  -$&- 
o-737 
=  746  watts. 

The  kilowatt  (kw.)  is  a  commonly  used  unit  of  power  and  is 
equal  to  1000  watts. 

7.  Resistance.  —  Resistance  is  the  inherent  property  of  a  ma- 
terial which  opposes  the  flow  of  an  electric  current,  and  by  virtue  of 
which  electrical  energy  is  converted  into  heat.     It  is  analogous,  in 
many  respects,  to  mechanical  friction,  and  has  been  experimen- 
tally determined  to  be   directly  proportional  to  length,  and  in- 
versely proportional  to  cross-sectional  area. 

d) 

area 

The  unit  of  length  commonly  used  is  the  foot;  that  of  area,  the 
circular  mil.  The  constant  p  is,  then,  the  resistance  in  ohms  of  a 
section  of  the  material  one  foot  long  and  having  a  cross-sectional 
area  of  one  circular  mil.  The  average  value  of  p  for  commercial 
copper  may  be  taken  as  10.8  at  ordinary  temperatures  (25°  C.). 

Materials  whose  resistivity  is  low,  such  as  copper,  silver,  alumi- 
num, etc.,  are  known  as  conductors;  those  whose  resistivity  is 
high,  such  as  glass,  rubber,  mica,  porcelain,  etc.,  are  known  as 
insulators. 

8.  Circular  mil.  —  The  circular  mil  is  an  arbitrary  unit  of  area 
much  used  in  electrical  calculations.     Since  electrical  conductors 
are  largely  circular  in  cross  section,  it  is  convenient  to  have  a  unit 
of  area  that  bears  a  simple  relation  to  the  diameter  of  a  circle.     The 
area  of  a  circular  cross  section,  in  circular  mils,  is  obtained  by  squar- 
ing the  diameter  in  thousandths  of  an  inch  (mils).     Hence,  the  area 
in  circular  mils  is  to  the  area  in  square  measure  as  4  is  to  TT. 

area  in  circular  mils 4  /  \ 

area  in  millionths  of  a  square  inch      TT 

9.  Temperature  coefficient.  —  An  increase  in  the  temperature 
of  a  copper  wire  causes  its  resistance  to  increase.     The  ratio  of 
the  change  in  the  resistance  of  a  conductor,  per  degree  change  in 


ESSENTIALS   OF  ELECTRICAL  ENGINEERING 


its  temperature,  to  its  resistance  at  its  initial  temperature,  is  its 
temperature  coefficient.  The  temperature  coefficient  of  most 
materials  is  not  constant,  but  changes  with  the  temperature.  By 
reference  to  Table  I  it  will  be  found  that  the  temperature  coeffi- 
cient of  copper  decreases  as  the  temperature  increases. 

TABLE   I 
TEMPERATURE  COEFFICIENTS  FOR  COPPER 

i 


a  = 


234-5  +  T 


T 

a 

T 

a 

r 

a 

T 

a 

0 

0.00427 

13 

o  .  00404 

26 

o  .  00383 

39 

o  .  00366 

i 

0.00425 

14 

o  .  00403 

27 

0.00381 

40 

o  .  00364 

2 

O  .  00424 

15 

o  .00401 

28 

o  .  00380 

4i 

0.00362 

3 

0.00422 

16 

o  .  00399 

29 

0.00379 

42 

0.00361 

4 

0.00420 

17 

0.00397 

30 

0.00378 

43 

o  .  00360 

5 

0.00418 

18 

o  .  00396 

31 

0.00377 

44 

o  .  00360 

6 

0.00416 

19 

o  .  00394 

32 

0.00375 

45 

0.00358 

7 

0.00414 

20 

0.00393 

33 

0.00374 

46 

0.00356 

8 

0.00412 

21 

0.00391 

34 

0.00373 

47 

0.00355 

9 

0.00411 

22 

o  .  00390 

35 

0.00371 

48 

0.00354 

10 

o  .  00409 

23 

o  .  00388 

36 

0.00370 

49 

0.00352 

ii 

o  .  00408 

24 

0.00387 

37 

o  .  00369 

5° 

0.00351 

I  2 

o  .  00406 

2  'C 

o  0038^ 

38 

o  00367 

0 

v^  •  ^wo^0 

o 

w  .  w^)  v^  i 

Rt  =  R(i  db  at). 

R   =  the  resistance  at  temperature  T  °C. 
Rt  =  the  resistance  at  temperature  (T  =t  /). 

a   =  the  temperature  coefficient  at  temperature  T  °C. 

/   =  the  change  in  temperature. 

The  resistance  of  some  materials,  of  which  carbon  is  an  example, 
decreases  as  the  temperature  increases.  Glass,  at  ordinary  tem- 
peratures, has  a  very  high  resistivity,  but  in  the  liquid  state  its  re- 
sistivity is  comparatively  low.  These  materials  are  said  to  have 
negative  temperature  coefficients.  Certain  metallic  alloys  have  a 
practically  constant  resistance  over  a  wide  range  of  temperature. 

10.  Ohm's  Law.  —  The  electromotive  force  required  to  overcome 
the  opposition  due  to  the  resistance  of  any  circuit  is  proportional  to  the 
current  flowing  in  the  circuit. 

When  the  electromotive  force  (e)  is  expressed  in  volts  and  the 
current  (i)  in  amperes,  the  proportionality  factor  (R)  is  in  ohms. 

e  =  Ri.  (3) 


THE   ELECTRIC    CIRCUIT  7 

The  experimental  fact  stated  above  is  the  first  fundamental  law 
of  the  electric  circuit  and  was  first  demonstrated  and  enunciated  by 
Dr.  G.  S.  Ohm.  It  applies  equally  to  continuous,  pulsating  and 
alternating  currents. 

ii.  Inductance.  —  That  property  of  a  body  which  tends  to  main- 
tain any  existing  state  or  condition  of  motion  of  the  body  is  termed 
its  inertia.  The  force  (/)  required  to  neutralize  the  effect  of  inertia 
is  proportional  to  the  rate  at  which  the  speed  or  velocity  of  the  body 
changes,  and  the  proportionality  factor  (M)  is  the  mass  of  the  body. 
When  the  velocity  of  a  body  is  changed  from  V'  to  V"  in  /  seconds 

,      M  (V  -  V") 
av./=-      —      ^,  (4) 

and  /  =  M  X  rate  at  which  V  changes.  (5) 

Suppose  the  speed  of  a  steam  engine  is  increased  from  100  r.p.m.  to 
200  r.p.m.  in  10  seconds.  A  force  is  required  to  overcome  the  inertia 
of  the  flywheel  and  other  rotating  parts,  but  the  energy  used  in 
causing  the  speed  of  the  engine  to  increase  is  stored  as  kinetic  energy 
in  the  rotating  parts,  and  is  returned  to  the  system  when  the  engine 
speed  decreases  to  its  original  value.  Therefore,  the  net  expendi- 
ture of  energy  due  to  the  inertia  of  the  moving  parts  of  a  machine  is 
zero  when  the  final  state  of  motion  is  the  same  as  the  initial  state. 

An  electric  circuit  has  a  property,  similar  to  that  of  inertia,  which 
tends  to  maintain  any  state  or  condition  of  current  flowing  in  the 
circuit.  The  electromotive  force  (e)  required  to  overcome  the  effect 
of  this  property  and  cause  the  current  to  increase  or  decrease,  is 
proportional  .  to  the  rate  at  which  the  current  changes,  and  the 
proportionality  factor  (L)  is  the  inductance  of  the  circuit.  The 
unit  of  inductance  is  the  henry.  When  the  current  in  a  circuit  is 
changed  from  if  to  i"  in  /  seconds 


T          —    >  /,-\ 

av.|0  =  L  -  j  (6) 

t 

and  e  —  L  X  rate  at  which  i  changes.  (7) 

It  is  evident  that  the  energy  expended  in  causing  a  current  to 
increase  is  equal  and  opposite  to  that  expended  in  causing  it  to 
decrease  to  its  original  value.  Therefore,  the  net  expenditure  of 
energy  due  to  the  inductance  of  an  electric  circuit  is  zero  when  the 
final  current  is  equal  to  the  initial  current. 


8 


ESSENTIALS  OF   ELECTRICAL  ENGINEERING 


12.   Capacitance.  —  In  the  system  shown  in  Fig.  4,  consisting  of  a 
pump,  pipe  connections,  and  a  cylinder  across  which  is  stretched  an 
elastic  membrane,  the  pressure  set  up  by  the  pump  stresses  the 
membrane  until  the   reaction   due  to   the 
stress  is  equal  to  the  pressure  of  the  pump, 
and    the   quantity   of   water    displaced    or 
stored  is  proportional  to  the  pressure. 

gallons  =  constant  X  pressure.         (8) 
FIG.  4.    Mechanical  Anal-  If  the  pressure  due  to  the  pump  decreases, 

ogy  of  Capacitance.  u  iT  i 

the  stress  in  the  membrane  becomes  less,  and 

energy  which  tends  to  maintain  the  original  pressure  set  up  by  the 
pump,  is  returned  to  the  system. 

An  electric  condenser  consists,  essentially,  of  two  conductors 
separated  by  an  insulating  material,  or  dielectric.  In  the  system 
shown  in  Fig.  5,  the  generator  sets  up 
a  pressure  which  stresses  the  dielectric 
separating  the  plates  PP  until  the 
reaction  equals  the  electromotive  force 
of  the  generator,  and  the  quantity  of 
electricity  displaced  or  stored  is  pro-  FlG*  5'  Electric  CaPacitance- 
portional  to  the  electromotive  force  (e)  of  the  generator.  The 
proportionality  factor  (C)  is  termed  the  capacitance  of  the  con- 
denser. The  unit  of  capacitance,  is  the  farad. 

q  =  Ce.  (9) 

When  the  applied  electromotive  force  is  changed  from  e'  to  e"  in 
/  seconds 

,e'-e" 


Resistance 


av.  i  =  C 


do) 


and 


i  =  C  X  rate  at  which  e  changes. 


If  the  electromotive  force  of  the  generator  decreases,  energy  is  re- 
turned to  the  system  and  tends  to  maintain  the  electromotive  force 
of  the  system  at  its  former  value,  the  stress  in  the  dielectric  being 
reduced  proportionately.  The  energy  returned  to  the  system 
when  the  condenser  is  completely  discharged  is  equal  to  the  energy 
stored,  the  condenser  being  so  constructed  that  the  losses  are 
negligible. 


THE  ELECTRIC   CIRCUIT  9 

13.  Alternating-current  circuits  containing  resistance  only.  — 

From  equation  (3) 

e  =  Ri  (12) 

=  RIm  smut,  (13) 

i.e.,  the  current  and  the  electromotive  force  in  an  alternating- 
current  circuit  containing  resistance  only,  are  in  phase  as  shown  in 
Fig.  6a. 

14.  Alternating-current  circuits  containing  inductance  only.  - 
Since  the  electromotive  force  required  to  overcome  the  effects  of 
inductance  is  proportional  to  the  rate  at  which  the  current  changes, 
the  electromotive  force  required  to  overcome  the  inductance  of  an 
alternating-current  circuit  in  which  the 

current  varies  harmonically  is  <°>    ' 

e 

e  =  uLIm  sin  («/  +  90°),*         (14) 


i.e.,  the  current  and  the  electromotive    ~|*' 


CO 


force  in  an  alternating-current  circuit 

containing   inductance    only,   are    90  Fl£.  <*'   Vectors Pof  Current  Mand 

Electromotive  Force  in  a  Non- 

degrees  out  of  phase,  and  the  current      inductive  Circuit. 

lags  behind  the  applied  electromotive  FIG.  6b.    Vectors  of  Current  and 

force  as  shown  in  Fig.  6b.  Electromotive   Force  in  an  In- 

.  i         f    .     /    .  0\       ductive  Circuit. 

The  maximum  value  of  sin  («/  +  90  )  FlG>  6c>    Vectors  o£  Current  and 
equals  I ,  and  Electromotive  Force  in  a  Capaci- 

Em  =  wLIm.  (15)       tive  Circuit. 

The  quantity  col,  is  termed  the  inductive  reactance  of  an  alter- 
nating-current circuit  and  is  expressed  in  ohms.     (Symbol  XL.) 

15.  Alternating-current  circuits  containing  capacitance  only. — 
Since  the  quantity  of  electricity  displaced  in  a  condenser  circuit  is 
proportional  to  the  applied  electromotive  force,  the  current  (rate 
of  displacement)  in  the  circuit  is  proportional  to  the  rate  of  change 
in  the  electromotive  force.  When  the  electromotive  force  varies 
harmonically 

i  =  uCEm  sin  (co/  +  90°),*  (16) 

i.e.,  the  current  and  the  electromotive  force  in  an  alternating-current 

circuit  containing  capacitance  only,  are  90  degrees  out  of  phase,  and 

the  current  leads  the  applied  electromotive  force  as  shown  in  Fig.  6c. 

*  See  Appendix  A,  Section  7. 


10  ESSENTIALS   OF  ELECTRICAL  ENGINEERING 

The  maximum  value  of  sin  («/  +  90°)  equals  i, 

Im  =  uCEm  '  (17) 

and  •;  IS.-f*.  ^  ||  (I8) 

The  quantity  —  is  termed  the  capacitive  reactance  *  of  an  alter- 
coC 

na ting-current  circuit  and  is  expressed  in  ohms.     (Symbol  Xc*) 

16.  Reactance  of  alternating-current  circuits.  —  The  reactance 
of  an  alternating-current  circuit  is,  from  equations  (15)  and  (18), 
the  ratio  of  the  current  flowing  in  the  circuit  and  the  electromotive 
force  required  to  overcome  the  combined  effects  of  inductance  and 
capacitance. 

17.  Alternating-current  circuits  containing  resistance   and  in- 
ductance in  series.  —  In  a  series  circuit  containing  resistance  and 
inductance,  the  applied  voltage  required  to  cause  a  given  current 
to  flow  in  the  circuit  must  be  equal  to  that  required  to  overcome 
the  combined  effects  of  resistance  and  inductance. 

e  =  eR  +  eL  (19) 

=  Ri  +  uLi  (20) 

=  RIm  sin  ut  +  (tiLIm  sin  («/  +  90°).  (21) 

The  right-hand  member  of  equation  (21)  is  the  surri  of  two  harmonic 
electromotive  forces  90  degrees  out  of  phase.  The  applied  electro- 
motive force  is,  therefore,  a  harmonic  quantity,  the  maximum  value 
of  which  is  the  geometric  sum  of  the  maximum  values  of  the  quad- 
rature electromotive  forces. 


Em    =  (RIm)2  +    (uLIm)2  (22) 


=  Im  VR2  +    (COL)2  (23) 

and  the  current  lags  behind  the  applied  electromotive  force  by  the 
angle 

0  =  tan-1—  (24) 

*  Capacitive  reactance  is  to  be  considered  a  negative  quantity. 


THE  ELECTRIC   CIRCUIT  II 

1 8.  Alternating-current  circuits  containing  resistance  and  ca- 
pacitance in  series.  —  In  a  series  circuit  containing  resistance  and 
capacitance, 

e  =  eR  +  ec  (25) 

=  Ri  +  ~dC  (26) 

r>  T  i   Im  sin  (ait  —  00°) .  /     \ 

=  RIm  sin  co/  +  -  •*  5  (27) 

therefore,  

"0*  (28) 


and  the  current  leads  the  applied  electromotive  force  by  the  angle 

(30) 


19.  Alternating-current  circuits  containing  resistance,  induc- 
tance and  capacitance  in  series.  —  In  a  series  circuit  containing 
resistance,  inductance  and  capacitance, 

e  =  eR  +  eL  +  ec  (31) 

=  Ri  -f  coLi  +  ^  (32) 

coC 

=  RIm  sin  «f  +  «L7,  sin  («/  +  90°)  +  /m  sin  (~  9°°).     (33) 


From  equation  (33)  Em  =  y  jR2/w2  +  («L/m  +  ^)'  (34) 


=  7m     ^  +   coL  +  (35) 


and  the  current  leads  or  lags  behind  the  applied  electromotive 
force,  as  the  quanti 
of  lead  or  lag  being 


force,  as  the  quantity  f  w£  +  ~;J  is  negative  or  positive,  the  angle 


XV 


(36) 


12  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

20.  Voltage   triangle.  —  When  an  alternating  current  flows  in 
a  circuit  containing  resistance  and  reactance  in  series,  the  applied 
electromotive  force  is  the  resultant  of  quadrature  electromotive 
forces,  and  the  voltages  of  the  circuit  may  be  represented  graphi- 
cally by  means  of   a  right  triangle. 
Fig.  ya.    Because  of  the  impossibility 
of  eliminating  resistance  from  a  reac- 
tance, just    as   it    is    impossible   to 
FIG.  7.    Voltage  Triangles.  .  ... 

eliminate    friction    from    a    machine 

having  moving  parts,  the  more  common  form  of  the  voltage  tri- 
angle is  an  obtuse  triangle,  the  hypotenuse  and  the  sides  of  which 
are,  respectively,  the  applied  electromotive  force,  the  voltage  be- 
tween the  terminals  of  the  resistor,  and  that  between  the  terminals 
of  the  reactor.  Fig.  yb. 
"N  21.  Impedance  and  the  impedance  triangle.  —  Dividing  equation 

(35)  by  Im}  and  representing  the  quantity  coL  -\ — —  by  X 

coC 

f?  =  VFTT2.  (37) 

J-m 


The  quantity  VR2  -f  X2  is  termed  the  impedance  of  an  alternating- 
current  circuit,  and  is  the  ratio  of  the  applied  electromotive  force 
and  the  current  flowing  in  the  circuit.  The 
unit  of  impedance  (Z)  is  the  ohm. 

The  impedance,  the  resistance  and  the  reac- 
tance    of    an    alternating-current   circuit   form    FlG- 8-    Impedance 
respectively,  the  hypotenuse,  the  base,  and  the 
altitude  of  a  right  triangle,  and  may  be  represented  graphically  as 
in  Fig.  8. 

Example.  —  A  series  circuit  has  the  following: 

Em  =    220  VOltS. 

/  =    60  cycles. 
R  =    30  ohms. 
L  =      o.i  henry. 
C  =      0.00013  farad. 
Find: 

(a)  The  reactance;  (b)  the  impedance;  (c)  the  current. 


THE  ELECTRIC   CIRCUIT  13 

Solution. 

XL  =  377  X  o.i  =  37.7  ohms. 

=  20.4  ohms. 


377  X  0.00013 
X  =  XL  -f-  Xc  =  17.3  ohms. 

Z  =  V(3o)2  +  (i7-3)2  =  34-7  ohms. 

r       Em      220 

Im  =  —  -  =  --  =  6.3  amperes. 

^       34-7 

22.  Alternating-current  circuits  containing  resistance  and  induc- 
tance in  parallel.  —  At  any  instant  the  total  current  supplied  to  a 
parallel  system  is  the  algebraic  sum  of  the  currents  flowing  in  the 
branches. 

i  =  *B  +  *i.  (38) 

By  definition  iR  =  (/m)^sinoj/.  (39) 

From  equation  (14) 

IL  =  (Im)L  sin  (co/  -  90°).  (40) 

Substituting  in  equation  (38), 

*  =  (Jm)fl  sin  w/  +  (7m)L  sin  («/  -  90°).  (41) 

The  right-hand  member  of  equation  (41)  is  the  sum  of  two  -harmonic 
currents  90  degrees  out  of  phase.  The  total  current  supplied  to  the 
system  is,  therefore,  a  harmonic  quantity,  the  maximum  value  of 
which  is  the  geometric  sum  of  the  maximum  values  of  the  currents 
flowing  in  the  branches. 

/.  =  \/CW+(/~)L2  (42) 

and  the  total  current  lags  behind  the  applied  electromotive  force 
by  the  angle 

=k.  (43) 


23.  Alternating-current  circuits  containing  resistance  and  capac- 
itance in  parallel.  —  When  resistance  and  capacitance  are  con- 
nected in  parallel 

*  =  **  +  «c>  (44) 

*«  =  (/•.)*  sin  «/.  (45) 


14  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

From  equation  (16) 

ic  =  (Im)c  sin  (co/  -f  90°).  (46) 

Substituting  in  equation  (44), 

i  =  (Im)R  sin  co/  +  (7m)c  sin  (co/  +  90°),  (47) 

Im    =   V(Im)R2  +   (Im)C2  (48) 

and  the  total  current  leads  the  applied  electromotive  force  by  the 
angle 

"T^T-  (49) 


24.  Alternating-current  circuits  containing  resistance,  induc- 
tance and  capacitance  in  parallel.  —  When  resistance,  inductance 
and  capacitance  are  connected  in  parallel 

*  =  iR  +  iL  +  ic,  (5°) 

**  =  (£»)«  sin  co/,  (51) 

iL  =  (7m)z,  sin  (co/  -  90°),  (52) 

ic  =  (Im)c  sin  (co/  +  90°).  (53) 

Substituting  in  equation  (50), 

i  =  (I™)R  sin  co/  +  (Im)L  sin  (co/  -  90°)  +  (7m)c  sin  (co/  +  90°),     (54) 

fi2  +  [(7»)z,  ~  (^m)cJ2  (55) 


and  the  total  current  leads  the  applied  electromotive  force  if 
[(7m)L  —  (7m) c]  is  negative,  and  lags  behind  the  applied  electro- 
motive force  if  [(7m)z,  —  (Im)c\  is  posi- 
tive,  the  angle  of  lead  or  lag  being 


tm^       VJ  (56) 

FIG.  9.     Current  Triangles.  {***)& 

25.  Current  triangle.  —  The  current  relations  of  a  parallel  cir- 
cuit are  shown  graphically  by  means  of  a  right  triangle.  Fig.  9a. 
Because  of  resistance  in  the  inductive  or  capacitive  branch  the 
currents  in  parallel  circuits  are  seldom  90  degrees  out  of  phase, 
and  Fig.  9b  shows  the  usual  relations  of  the  component  and  the 
resultant  currents. 


THE   ELECTRIC   CIRCUIT  15 

26.  Joule's  Law.  —  The  work  done  in  any  electrical  circuit  is 
proportional  to  the  square  of  the  current  flowing  in  the  circuit  and  to  the 
time  the  flow  continues. 

When  the  electromotive  force  (e)  is  expressed  in  volts,  the  current 
(i)  in  amperes  and  the  time  (/)  in  seconds,  the  unit  of  work  is  the 
joule  (A),  and  the  proportionality  factor  is  in  ohms. 

A  oc  «.  (57) 

Joule's  Law  is  the  second  fundamental  law  of  the  electric  circuit 
and,  like  Ohm's  Law,  has  been  shown,  experimentally,  to  be  uni- 
versal. 

27.  Work  done  in  an  electric  circuit.  —  From  equation  (57) 

AR  =  Ri^t  (58) 

when  current  flows  in  a  circuit  containing  resistance  only; 

AL  =  coL*2/  (59) 

when  current  flows  in  a  circuit  containing  inductance  only; 

Ac=^  '  (60) 

when  current  flows  in  a  circuit  containing  capacitance  only. 

28.  Power  in  an  electric  circuit.  —  Power  is,  by  definition,  the 
rate  of  doing  work.     Therefore, 

work  ,,  ^ 

Power  =  -T-  (61) 

time 

and  pR  =  RP  watts  (62) 

when  current  flows  in  a  circuit  containing  resistance  only; 

pL  =  uL?  watts  (63) 

when  current  flows  in  a  circuit  containing  inductance  only; 

*2 

pc  =  -7;  watts  (64) 

o?C 

when  current  flows  in  a  circuit  containing  capacitance  only. 
From  equation  (3) 

Ri  =  eR.  (65) 

From  equation  (14) 

uLi  =  CL.  (66) 


1  6  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

From  equation  (16) 


(67) 


Therefore,  the  instantaneous  power  (watts)  in  any  electrical  cir- 
cuit is  equal  to  the  product  of  the  applied  electromotive  force 
(volts)  and  the  current  (amperes)  flowing  in  the  circuit. 

p  =  ei.  (68) 
In  a  continuous-current  circuit,  e  and  i  are  constant  and 

P  =  EI  watts.  (69) 
In  an  alternating-current  circuit  containing  resistance  only 

pR  =  Em  sin.  utlm  sin  co/  (70) 

=  Emlm  sin2  co/.  (71) 

From  Appendix  A,  Section  9,  the  average  value  of  sin2  co/  during  one 
complete  cycle  is  ^.  Therefore, 

av.  pR  =  ^^  watts.  (72) 

2 

In  an  alternating-current  circuit  containing  inductance  only 

pL  =  Em  sin  co//m  sin  (co/  -  90°)  (73) 

=  Emlm  sin  co/  cos  co/,  (74) 

but  the  average  value  of  sin  co/  or  cos  co/  during  one  complete  cycle 
is  zero.*  Therefore, 

av.  pL  =  o.  (75) 

In  an  alternating-current  circuit  containing  capacitance  only 

pc  =  Em  sin  (co/  +  90°)  co//w  sin  co/  (76) 

=  Emlm  cos  co/  sin  co/,  (77) 
which  is  identical  with  equation  (74)  and 

av.  pc  =  o.  (78) 
In  any  alternating-current  circuit 

p  =  Em  sin  co/  Im  (sin  co/  ±  0)  (79) 

=  Em/w  sin  co/  (sin  co/  cos  0  =b  cos  co/  sin  0)  (80) 

=  Emlm  (sin2  co/  cos  0  ±  sin  co/  cos  co/  sin  0).  (81) 
*  See  Appendix  A,  Section  8. 


THE  ELECTRIC   CIRCUIT  17 

From  Appendix  A,  Section  9, 

av.  sin2  co/  =  \  (82) 

and 

av.  sin  co/  cos  co/  sin  <f>  =  o.  (83) 

Therefore, 

Emlm  cos  0  ,    ^ 

ay.  p  =  -  -  watts,  (84) 

i.e.,  the  average  power  *  (watts)  in  any  alternating-current  circuit 
is  equal  to  one-half  the  product  of  the  maximum  electromotive 
force  (volts),  the  maximum  current  (amperes)  and  the  cosine  of 
the  phase  angle. 

29.  Effective  current  and  electromotive  force.  —  The  steady 
current  (or  electromotive  force)  which,  acting  in  a  circuit  of  constant 
resistance  and  zero  reactance,  transfers  energy  at  the  average  rate  of 
transfer  when  an  alternating  current  (or  electromotive  force)  acts  in 
the  same  circuit,  is  the  elective  value  of  the  alternating  current  (or 
electromotive  force)  . 

From  equation  (62) 

P  =  R?  (85) 

=  RIm2sm2wt.  (86) 

Then  av.  p  =  RImz  av.  (sin2  «/)t  /  (87) 

=  ~-  (88) 

For  continuous  currents 

P=RI2.  (89) 

Therefore, 

/?/"  2 
RP  =  &*-  (go) 

and 


2  _j 

=  0.707  Im.^  (92) 

Similarly, 

^  =  ^ 
R       2  R 

X 

*  A  watt-meter  indicates  the  average  power  in  an  alternating-current  circuit. 
t  See  Appendix  A,  Section  9. 


1 8  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

and 

£  =  ^  (94) 

V2 
=  0.707  Em.  (95) 

Therefore,  the  effective  value  of  a  harmonic  alternating  current  (or 
electromotive  force)  is  equal  to  its  maximum  value  divided  by  the 
square  root  of  2,  and  effective  values  may  be  substituted  in  any  of 
the  above  equations  containing  maximum  values. 

Effective  values  are  indicated  by  ammeters  and  voltmeters,  and 
are  always  to  be  assumed  in  alternating-current  problems  unless  the 
values  given  or  required  are  specifically  stated  to  be  maximum  or 
instantaneous. 

30.  Power  factor.  —  The  cosine  of  the  phase  angle  (cos  0)  is 
termed  the  power  factor  of  an  alternating-current  circuit.     The 
power  factor  may  also  be  expressed  as  the  ratio  of  the  resistance  to 
the  impedance, 

•n 

COS  0  =  ~  ,  (96) 

or  by  the  ratio  of  the  watts  to  the  product  of  volts  and  amperes, 

cos0  =  |^.  (97) 

31.  Resistances  in  series.  —  The  resistance  of  a  series  circuit  is 
the  arithmetical  sum  of  the  resistances  of  its  parts. 

E  =  Ei  +  E2  (98) 

=  7  (^  +  £2).  (99) 

Dividing  by  I 

R  =  £1  +  Ri.  (100) 

32.  Reactances  in  series.  —  The  reactance  of  a  series  circuit  is 
the  algebraic  sum  of  the  reactances  of  its  parts. 

E  =  Ei  4-  E2  (101) 

=  /(X1  +  Z2).y  (102) 

Dividing  by  7 

X  =  Xi  +  X2.  (103) 


THE   ELECTRIC    CIRCUIT 


33.   Impedances  in  series.  —  The  impedance  of  a  series  circuit 
is  the  geometric  sum  of  the  impedances  of  its  parts.     From  equation 

(35) 


E  =      Er2  +  Ex*  (104) 

=  /  y/(Ri  +  Rtf  +  (Xl  +  X2y.  (105) 

Dividing  by  / 


Z  =       (R,  +  R2)*  +  (Xi  +  X2)2.  (106) 

34.  Resistances  in  parallel.  —  In  a  parallel  system  the  branches 
of  which  contain  resistance  only,  the  reciprocal  of  the  resistance  of 
the  system  is  the  arithmetical  sum  of  the  reciprocals  of  the  resist- 
ances of  the  branches. 

/  =  /I  +  /2,  (I07) 


Dividing  by  E 


35-  Reactances  in  parallel.  —  In  a  parallel  system  the  branches 
of  which  contain  reactance  only,  the  reciprocal  of  the  reactance  of 
the  system  is  the  algebraic  sum  of  the  reciprocals  of  the  reactances 
of  the  branches. 

/  =  /l+/2,  •        (HO) 

-  =  —  +—. 

Dividing  by  E 


36.  Resistance  and  reactance  in  parallel.  —  In  a  parallel  system 
consisting  of  a  branch  containing  resistance  only  and  a  branch  con- 
taining reactance  only,  the  reciprocal  of  the  impedance  of  the 
system  is  the  geometric  sum  of  the  reciprocal  of  the  resistance  and 
the  reciprocal  of  the  reactance  of  the  branches. 

From  equation  (55) 

/  =   V/!2  +  /22,  (113) 

E 


/E\2  ,    ^ 

(x)  •  (I14) 


20  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

Dividing  by  E 

'~  f^r       I      .         ("5) 

37.   Impedances  in  parallel.  —  Any  impedance  containing  both 
resistance  and  reactance  (Fig.  10)  may  be  replaced  by  a  parallel 


FIG.  10.     Resistance  and  Inductance  FIG.  n.     Resistance  and  Inductance 

in  Series.  in  Parallel. 

system  the  branches  of  which  contain  only  resistance  or  reactance 
(Fig.  n).  Referring  to  Fig.  10  let 

E  =  the  applied  electromotive  force, 
/  =  the  current, 
Z  =  the  impedance, 
cos  0  =  the  power  factor. 

Then       Ir  =  I  cos  </>  (=  power  component  of  current), 
Ix  =  I  sin  </>  ( =  wattless  component  of  current), 
Rs  =  Z  cos  <f>, 
X8  =  Z  sin  4>. 

Referring  to  Fig.  n,  let 

E  =  the  applied  electromotive  force, 

/  =  the  line  current, 

Ir  —  the  current  in  the  resistance, 

Ix  =  the  current  in  the  inductance, 
Rp  =  the  resistance, 
Xp  =  the  reactance, 

Z  =  the  impedance. 

i.e.,  the  applied  electromotive  force,  the  current,  the  impedance  and 
the  power  factor  of  one  system  are  made  equal  to  those  of  the  other 
system,  and  the  systems  are,  therefore,  equivalent. 

R,  =  f-  (»6) 

•*•  r 

(it?) 


COS 


Z  cos 
Z2 


ELECTRIC   CIRCUIT  2I 

71 

(119) 

(»o) 
(iai) 

(122) 

(123) 
(124) 
(125) 


sin  <£ 

Z2 

Z  sin<£ 

Z2 


From  equation  (115) 

'\  +  ^k  (126) 


(127) 

Therefore,  for  a  parallel  system, 


Z  V     VZl2     '       7  2      '      ^  2/        r    ^  2  "T    'Z  2  +    ^"J   »  (I2&) 


I?  7-2  /  -^1       I     -^2       ,     -«Vn    \  /  \ 

^  =  Z  l^  +  ^l  +  ^l)'  (I29) 

(130) 


Example.  —  Let  the  circuit  shown  in  Fig.  12  have:  RI  =  6  ohms, 
i?2  =  8  ohms,  J?3  =  10  ohms,  J¥"i  =  5  ohms,  J\T2  =  4  ohms,  -Xs  =  3  ohms. 
Then, 


Z22  =    82  +  42  =  80,  |x-    f-   |x' 

Z32  =  io2  +  ^2  =  100,  FlG-  I2'    Impedances 

in  ParalleL 


22  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

and 


=  v++          +    + 

Z       '  \6i      80      ice/  \6i      80      109 

=  V(0.098  +0.1+0.092)2  +  (0.082  +  0.05  +  0.028)2 
=  0.331. 

Z  =  -     -  =  3.02  ohms. 

R  =  (3-02)2  (0.098  +  o.i  +  0.092)  =  2.65  ohms, 
X  =  (3.02)2  (0.082  +  0.05  +  0.028)  =  1.46  ohms. 
Also, 

tan  0  =  0-082+0.05,+ 0.028  =  0.159  =  Q 

0.098  +  o.i  +  0.092       0.289 
cos  0  =  0.876. 

R  =  Z  cos  0  =  3.02  X  0.876  =  2.65  ohms. 

sin  0  =  0.482. 
X  =  Z  sin  0  =  3.02  X  0.482  =  1.46  ohms. 

38.  Resonance  of  alternating-current  circuits.  —  In  a  series 
circuit  the  electromotive  force  required  to  overcome  the  effect  of 
inductance  leads  the  current  by  90  degrees ;  the  electromotive  force 
required  to  overcome  the  effect  of  capacitance  lags  behind  the  cur- 
rent by  90  degrees.  These  two  harmonic  forces  are,  then,  180 
degrees  out  of  phase,  and  tend  to  neutralize.  When  the  inductive 
electromotive  force  is  equal,  numerically,  to  the  capacitive  elec- 
tromotive force,  the  circuit  is  said  to  have  voltage  resonance. 

I 


=o 


(131) 


Example.  —  Find  the  capacitance  required  to  neutralize  an  in- 
ductance of  o.i  henry  in  a  6o-cycle  alternating-current  circuit, 
the  capacitance  to  be  connected  in  series  with  the  inductance. 

Solution.  —  From  equation  (131) 

r_  _!_ 
*•/  —    nT 


7T2  X  4  X  3600  X  o.i 
=  0.007042  farad. 

In  a  parallel  circuit,  the  wattless  component  of  current  in  an 
inductive  branch  lags  90  degrees  behind  the  applied  electromotive 


THE   ELECTRIC   CIRCUIT  23 

force;  the  wattless  component  of  current  in  a  capacitive  branch 
is  90  degrees  ahead  of  the  applied  electromotive  force.  These 
wattless  components  of  current  are,  then,  180  degrees  out  of  phase, 
and  tend  to  neutralize.  When  the  leading  wattless  component 
is  equal,  numerically,  to  the  lagging  wattless  component,  the  cir- 
cuit is  said  to  have  current  resonance. 

/iag  sin  0'  =  /iead  sin  0"  (132) 

when      /iag  =  the  current  flowing  in  the  inductive  branch, 
/lead  =  the  current  flowing  in  the  capacitive  branch, 
4>'  =  the  angle  between  the  vector  of  electromotive  force 

and  that  of  the  current  in  the  inductive  branch, 
0"=  the  angle  between  the  vector  of  electromotive  force 
and  that  of  the  current  in  the  capacitive  branch. 

Example.  —  The  impedance  of  a  6o-cycle  inductive  circuit  is 
22  ohms,  the  current  flowing  in  the  circuit  is  10  amperes,  and 
lags  30  degrees  behind  the  applied  electromotive  force.  Find  the 
capacitance  to  be  connected  in  parallel  with  the  impedance,  the 
electromotive  force  and  the  resultant  current  to  be  in  phase. 

Solution.  —  The  voltage  across  the  condenser  is 

EC  =  22  X  10  =  220  VOltS, 

and  the  wattless  component  of  current  is 

Iw  =  10  X  0.5  =  5  amperes. 
From  equation  (18) 

C  = s 

377  X  220 

=  0.00006  farad. 

39.  Use  of  electrical  measuring  instruments.  —  A  voltmeter  is 
connected  in  parallel  with  that  part  of  a  circuit  in  which  it  is  desired 
to  measure  the  drop  of  poten- 
tial. Fig.  1 3  a. 

An  ammeter  is  connected        ^—^\  <&> 

in  series  with  the  circuit  in 
which  it  is  desired  to  measure     FIG.  13.     Voltmeter,  Ammeter  and  Watt- 
the  current,  the  entire  cur- 

rent,  or  a  constant  proportion  of  it,  going  through  the  coils  of  the 
instrument.  Fig.  i3b. 

A  wattmeter  consists  of  two  current-carrying  coils,  a  voltage 


24  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

coil  connected  in  parallel  with  the  load  and  a  current  coil  con- 
nected in  series  with  the  load.  The  reaction  set  up  by  the  coils  is 
proportional  to  the  product  of  the  currents  flowing  in  the  coils,  i.e., 
to  the  power  in  the  circuit,  and  causes  the  deflection  of  a  movable 
element. 

Permanent-magnet  instruments  indicate  on  continuous-current 
circuits  only;  induction  instruments  on  alternating-current  circuits 
only;  dynamometer  instruments  on  either  continuous-  or  alternat- 
ing-current circuits. 

CHAPTER  I  — PROBLEMS 

1.  Find  the  area  in  circular  mils  of:  (a)  a  circle  \  inch  in  diameter,  (b)  a  rec- 
tangle i  X  i  inch. 

2.  Find  the  diameter  of  a  circle,  the  area  of  which  is:  (a)  211,600  circular 
mils,  (6)  52,240  circular  mils,  (c)  10,400  circular  mils,  (d)  1000  circular  mils. 

3.  Find  the  side  of  the  square,  the  area  of  which  is:  (a)  211,600  circular 
mils,  (b)  52,240  circular  mils,  (c)  10,400  circular  mils,  (d)  1000  circular  mils, 

4.  Find  the  resistance,  at  20°  C.,  of  1000  feet  of  copper  wire:  (a)  o.i  inch  in 
diameter,  (b)  0.25  inch  in  diameter,  (c)  0.6  inch  in  diameter,  (d)  i  inch  in 
diameter. 

5.  Find  the  resistances  of  the  wires  specified  in  Problem  4  at  a  temperature 
of  45°  C. 

6.  Find  the  electromotive  force  required  to  cause  a  current  of  10  amperes 
to  flow  in  a  circuit,  the  resistance  of  which  is:  (a)  i  ohm,  (b)  5  ohms,  (c)  18 
ohms,  (d)  25  ohms,  (e)  40  ohms. 

7.  Find  the  average  electromotive  force  of  self-induction  in  a  circuit,  the 
inductance  of  which  is  i  henry,  when  the  current  changes:  (a)  from  i  ampere  to 
5  amperes  in  o.i  of  a  second,  (b)  from  10  amperes  to  zero  in  o.oi  of  a  second,  (c) 
from  25  amperes  to  zero  in  0.2  second. 

8.  Find  the  capacitance  of  a  circuit  when  5  coulombs  of  electricity  are  dis- 
placed by  an  electromotive  force  which  changes:  (a)  from  25  volts  to  zero  in  i 
second,  (b}  from  50  volts  to  100  volts  in  \  of  a  second,  (c)  from  zero  to  10  volts 
in  o.oi  of  a  second. 

9.  The  resistance  of  an  electric  circuit  is  10  ohms.     Find  the  maximum 
value  of  the  current  when  the  maximum  value  of  the  alternating  electromotive 
force  applied  to  the  terminals  of  the  circuit  is:  (a)  25  volts,  (b)  40  volts,  (c)  90 
volts,  (d)  150  volts,  (e)  500  volts. 

10.  The  inductance  of  a  circuit  is  o.i  henry.     Find  the  electromotive  force 
required  to  cause  an  alternating  current  of  100  amperes  (maximum)  to  flow  in 
the  circuit  when  the  frequency  of  the  alternating  current  is:  (a)  10,  (b)  25,  (c) 
40,  (d)  50,  (e}  60,  (/)  100. 

11.  Find  the  voltage  required  to  charge  a  6o-microfarad  condenser  with  5 
coulombs  of  electricity. 

12.  The  charge  on  a  condenser  changes  from  i  coulomb  to  2  coulombs  in 


THE   ELECTRIC   CIRCUIT  25 

o.oi  of  a  second.     Find  the  average  current  flowing  in  the  circuit  during  the 
period  of  change. 

13.  An  electric  circuit  has  an  inductance  of  o.  i  henry.    '  Find  the  capacitance 
required  to  cause  resonance  when  the  frequency  of  the  alternating  electro- 
motive force  is:  (a)  25,  (b)  40,  (c)  60,  (d)  100. 

14.  Find  the  impedance  of  an  electric  circuit  having  a  resistance  of  8  ohms 
and  a  reactance  of  6  ohms  connected  in  series. 

15.  Find  the  reactance  of  a  circuit  having  an  inductance  of  0.15  henry 
when  the  frequency  of  the  applied  electromotive  force  is:  (a)  15,  (6)  25,  (c)  40, 
(d)  60,  (e)  100. 

16.  Find  the  reactance  of  a  lo-microfarad  condenser  when  the  frequency 
of  the  applied  electromotive  force  is:  (a)  15,  (b)  25,  (c)  40,  (d}  60,  (e)  100. 

17.  The  resistance  of  an  electric  circuit  is  2  ohms.     Find:  (a)  the  total  heat 
developed  in  the  circuit  when  50  amperes  continuous  current  flows  steadily  in 
the  circuit  for  10  minutes,  (b)  the  maximum  value  of  the  alternating-current 
required  to  develop  the  same  quantity  of  heat  in  the  same  length  of  time,  (c) 
the  power  (watts)  in  the  circuit. 

1 8.  The  continuous  electromotive  force  applied  to  an  electric  circuit  is  220 
volts.    This  electromotive  force  produces  a  current  of  20  amperes.     Find:  (a) 
the  resistance  of  the  circuit,  (b)  the  power  used  in  the  circuit. 

19.  The  effective  value  of  an  alternating  current  is  20  amperes,  the  applied 
electromotive  force  (effective)  is  220  volts,  and  the  power  factor  of  the  circuit 
is  0.85.   Find:  (a)  the  impedance  of  the  circuit,  (b)  the  resistance  of  the  circuit, 

(c)  the  reactance  of  the  circuit,  (d)  the  power  used  in  the  circuit. 

20.  A  rheostat  and  a  condenser  are  connected  in  series  to  a  no-volt,  25- 
cycle  alternating-current  generator.     The  value  of  the  resistance  is  10  ohms  and 
the  capacitance  of  the  condenser  is  o.i  microfarad.     Find:  (a)  the  reactance 
of  the  circuit,  (b)  the  impedance  of  the  circuit,  (c)  the  power  factor  of  the  circuit, 

(d)  the  current  flowing  in  the  circuit,  (e)  the  power  component  of  electromotive 
force,  (/)  the  wattless  component  of  electromotive  force. 

21.  Three  impedances,  the  values  and  power  factors  of  which  are  indicated 
below,  are  connected  in  series.     Find:  (a)  the  impedance  of  the  circuit,  (b)  the 
power  factor  of  the  circuit. 

Zi  =  10  ohms.  p.  f.  =  0.9  (lagging). 

Z2  =  15  ohms.  p.  f.  =  i. 

Z3  =  20  ohms.  p.  f.  =  0.6  (lagging). 

22.  The  impedances  specified  in  Problem  21  are  connected  in  parallel. 
Find:  (a)  the  impedance  of  the  circuit,  (b)  the  resistance  of  the  circuit,  (c)  the 
reactance  of  the  circuit,  (d)  the  power  factor  of  the  circuit. 

23.  Find  the  resistance  of  1000  feet  of  copper  wire,  102  mils  in  diameter  and 
at  a  temperature  of  40°  C. 

24.  Find  the  length  of  copper  wire,  204  mils  in  diameter,  the  resistance  of 
which,  at  the  same  temperature,  is  equal  to  the  resistance  of  the  wire  specified 
in  Problem  23. 

25.  The  resistance  of  the  field  windings  of  a  shunt  motor  was  found  to  be 
70  ohms  at  20°  C.    After  the  motor  had  been  in  operation  for  three  hours,  it  was 


26 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


found  that  the  resistance  of  the  windings  was  10  per  cent  greater  than  when 
the  first  measurement  was  made.     Find  the  temperature  of  the  field  windings. 

26.  Two  loads  A  and  B  are  connected  in  parallel  to  alternating-current 
mains.     The  power  factor  of  A  is  0.9  and  Ia  —  100  amperes;  the  power  factor 
of  B  is  0.3  and  Ib  —  50  amperes.     Find:  (a)  the  current  i;i  the  line,  (b)  the  power 
factor  of  the  system. 

27.  Measurements  taken  on  the  series  system  indicated  in  the  accompany- 

ing  sketch  were  as  follows: 

I  =  100  amperes,  E  =  2300  volts,  E\  =  1800  volts, 
E2  =  goo  volts. 


22(7  V. 

t 

—  200  Ft.  -*p  200  'Ft  >K-  ZOO  Ft  M 

1 

-X—  '  I 

flains 

:  | 

1 

J215  Volts 

1 

| 

25Amp& 

-i-b 

2/5  Vo//s 
15  Amperes 

2/5  V&//S 
lOAmperes 

Find:  (a)  the  power  factor  of  the  circuit,  (b)  the  resistance  of  the  circuit,  (c)  the 
reactance  of  the  circuit,  (d)  the  impedance  of  the  circuit,  (e)  the  power  used  in 
the  circuit. 

28.  The  continuous-current  distributing  system  indicated  in  the  accom- 
panying sketch  is  connected  to  2  20-  volt  supply  mains.  Determine:  (a)  the  sizes 
of  wires  required  for  the  dif- 
ferent parts  of  the  circuit  so 
that  215  volts  may  be  de- 
livered at  the  terminals  of 
each  load,  (b)  the  resistance 
of  the  distributing  system  (not 
including  the  load),  (c)  the  power  (watts)  loss  due  to  the  resistance  of  the  wires, 

(d)  the  total  output  of  the  generator. 

R   pV\^-.  29.   Calculate  the  resistance  of  the  circuit  indicated  in 

—  VWH        -   the  accompanying  sketch  when: 

*-M/\r 
Rs  RI  =  10  ohms,  Rz  =5  ohms,  Rz  =  8  ohms. 

30.  Find  the  impedance  of  the  system  specified  in  Problem  29  when  a  re- 
actance of  2  ohms  is  connected  in  series  with  Rz. 

31.  The  capacitance  of  each  of  three  condensers  is  10  microfarads.     Find 
the  capacitance  of  the  circuit  when:  (a)  the  condensers  are  connected  in  series, 
(b)  the  condensers  are  connected  in  parallel. 

32.  Find  the  current  flowing  in  a  25-microfarad  condenser  when  connected 
to  a  2  20-  volt,  6o-cycle,  alternating-current  system. 

33.  Three  impedances  A  ,  B  and  C  are  connected  in  series  to  6o-cycle  mains. 

Ra  =  loohms,  Rb  =  5  ohms,  Rc  =  12  ohms,  La  —  o.i  henry, 
Lb  =  0.08  henry,  Lc  =  0.2  henry. 

A  current  of  10  amperes  (effective)  flows  in  the  circuit.  Find  the  voltage  be- 
tween the  terminals  of:  (a)  impedance^!,  (b)  impedance  B,  (c}  impedance  C,  (d~) 
the  circuit  (the  applied  voltage). 

34.  The  impedances  specified  in  Problem  33  are  connected  in  parallel  and 
to  no-  volt,  25-cycle  mains.     Find:  (a)  the  impedance  of  the  system,  (b)  the 
resistance  of  the  system,  (c)  the  reactance  of  the  system,  (d)  the  current  in 
each  parallel  branch,  (e}  the  total  current  flowing  in  the  circuit  (line)  . 


THE  ELECTRIC   CIRCUIT  27 

35.  An  induction  motor  is  operated  in  parallel  with  100  incandescent  lamps. 
The  motor  takes  50  amperes  and  has  a  power  factor  of  0.8;  the  lamps  each  re- 
quire 0.4  ampere  and  their  power  factor  is  unity.     Find:  (a)  the  current  sup- 
plied to  the  combination,  (b)  the  power  factor  of  the  circuit,  (c)  the  power 
component  of  line  current,  (d)  the  wattless  component  of  line  current. 

36.  The  current  in  a  line  supplying  two  induction  motors  is  100  amperes,  and 
lags  45  degrees  behind  the  electromotive  force.    One  motor  has  a  power  factor 
of  0.85,  and  takes  60  amperes.    Find  the  power  factor  of  the  other  motor,  and 
the  current  supplied  to  it. 

37.  The  maximum  value  of  an  alternating  current  flowing  in  a  circuit,  the 

impedance  of  which  is  25  ohms  and  the  ratio  —  =0.75,  is  100  amperes.     Find: 

JK. 

(a)  the  effective  value  of  the  current,  (b)  the  power  in  the  circuit,  (c)  the  power 
factor  of  the  circuit. 

38.  Find  the  size  wire  required  to  supply  10  amperes  to  lamps  200  feet  from 
the  generator,  when  the' allowable  resistance  of  the  wires  is  0.2  ohm. 

39.  Find  the  current  flowing  in  the  circuit  when  the  impedances  of  Problem 
33  are  connected  to  25-cycle  mains,  the  voltage  of  which  is  equal  to  the  voltage 
found  in  part  (d)  of  Problem  33. 

40.  Three  impedances  are  connected  in  parallel. 

fa  —  73  and  lags  30  degrees. behind  the  electromotive  force, 
Ib  =  150  and  lags  45  degrees  behind  the  electromotive  force, 
Ic  =  225  and  lags  15  degrees  behind  the  electromotive  force.     Find:  (a) 
the  current  in  the  supply  line,  (b)  the  power  factor  of  the  system. 

41.  The  resistances  and  reactances  indicated  in 
the  accompanying  sketch  have  the  following  values: 

Ri  =  5,  &  =  3,  ^s  =  6,  R4  =  10,  Xl  =  8, 
Xz  =  4,  X,  =  3. 

Find:  (a)  the  impedance  of  the  circuit,  (b)  the  power 
factor  of  the  circuit. 

T       42.   Referring  to  the  accompanying  sketch: 

EI  =  100,  £2  =  150,  /  =  10,  EI  and  Ez  are  60  degrees 

:  out  of  phase. 

E,     j    Find:  (a)  E,  (6)  Z,  (c)  R,  (d)  X,  (e)  power  factor,  (/) 
Li-  power  in  the  circuit. 


43.  Find  the  impedance,  the  resistance, 
the  reactance  and  the  power  factor  of  the 
following  circuit: 


44.   Find  the  impedance,  the  resistance  and  the  reactance  of  the  circuit  in 
Problem  43  when  Xi  is  a  capacitance  equivalent  to  5  ohms. 


28 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


45.   Referring  to  the  accompanying  figure: 

R  r 


E  =  220  volts,  R  =  40  ohms,  r  =  0.5  ohm,  L  =  o.i  henry, 
C  =  0.00012  farad,  /  =  60  cycles. 

Find:  (a)  the  current,  (&)  22i,  (c)  £2,  (d)  E3,  (e)  power  in  the  circuit,  (/)  power 
factor. 
Draw  vector  diagram  representing  the  current  and  the  voltages  in  the  circuit. 

46.  A  voltmeter,  the  resistance  of  which  is  3,000  ohms  (reactance  negligible), 
is  connected  in  series  with  a  condenser  between  25o-volt,  oo-cycle  alternating- 
current  mains.    The  indication  of  the  meter  is  1 50.     Find  the  capacitance  of 
the  condenser,  and  the  voltage  between  its  terminals. 

47.  A  current  of  8  amperes  flows  in  an  impedance  connected  between  240- 
volt,  6o-cycle  alternating-current  mains.     The  power  absorbed  in  the  imped- 
ance is  1280  watts.     Find:  (a)  the  power  factor  of  the  circuit,  (b)  the  resistance 
of  the  circuit,  (c)  the  reactance  of  the  circuit,  (d)  the  inductance  of  the  circuit. 
Draw  vector  diagram  of  the  voltages  and  the  current. 

48.  When  an  impedance  (resistance  and  inductance  in  series),  the  power 
factor  of  which  is  0.707,  is  connected  to  6o-cycle,  2 50- volt  alternating-current 
mains,  25  amperes  flow  in  the  circuit.     Find:    (a)  the  capacitance  required 
to  be  connected  in  the  circuit  to  cause  voltage  resonance,  (b)  the  current  in 
the  resonant  circuit,  (c)  the  voltage  across  the  resistance,  (d)  the  voltage 
across  the  inductance,  (e)  the  voltage  across  the  capacitance. 

49.  The  impedance  in  Problem  48  is  connected  to  6o-cycle,  25o-volt  alter- 
nating-current mains  in  parallel  with  a  condenser.     The  resistance  of  the  con- 
denser circuit  is  negligible.     Find  the  capacitance  when  the  power  factor  of 
the  circuit  is:   (a)  unity,  (b)  0.90  lagging,  (c)  o.oo  leading. 

50.  Same  as  Problem  49  except  the  resistance  of  the  condenser  circuit  is 
5  ohms. 

51.  Show  that  when  the  applied  voltage  and  the  resistance  of  a  series  cir- 
cuit are  constant,  and  the  reactance  of  the  circuit  varies  from  zero  to  infinity, 
the  locus  of  the  current  vector  is  a  semi-circle,  the  diameter  of  which  makes 
zero"  angle  with  the  vector  of  applied  electromotive  force. 

52.  Show  that  when  the  applied  voltage  and  the  reactance  of  a  series  cir- 
cuit are  constant,  and  the  resistance  of  the  circuit  varies  from  zero  to  infinity, 
the  locus  of  the  current  vector  is  a  semi-circle,  the  diameter  of  which  makes 
an  angle  of  90  degrees  with  the  vector  of  applied  electromotive  force. 


CHAPTER  II 
MAGNETISM   AND   MAGNETIC   INDUCTION 

1.  Magnetism.  —  A  body  which  has  the  power  to  attract  a  piece 
of  iron  is  termed  a  magnet,  and  the  property  by  virtue  of  which 
attraction  takes  place  is  termed  magnetism.     The  ultimate  nature 
of  magnetism,  like  that  of  electricity,  has  never  been  determined, 
but  electricity  and  magnetism  are  intimately  associated. 

2.  Magnetic  field.  —  The  space  in  and  around  a  magnet  is  a 
magnetic  field.     A  magnetic  field  is  represented  graphically  by 
means  of  lines  (lines  of  magnetic  force  or  magnetic  induction)  which 
pass  from  the  north  pole  to  the  south  pole  of  a 

magnet  and  form  complete  loops  or  circuits. 
Fig.  14.  The  total  number  of  lines  of  magnetic 
induction  leaving  a  north  pole  and  entering  a 
south  pole  is  a  magnetic  flux.  FlG-  X4- 

3.  Production  of  a  magnetic  field.  —  The  only  known  means  for 
the  production  of  a  magnetic  field  is  the  electric  current  (electricity 

in  motion).  It  has  been  proved  experimentally 
that  the  space  surrounding  a  current-carrying  con- 
ductor is  a  magnetic  field,  that  the  lines  of  mag- 
netic induction  are  concentric  circles,*  and  that 
their  direction  is  clockwise  or  counter-clockwise  as 
the  current  in  the  conductor  flows  away  from  or 
toward  the  observer.  The  direction  of  a  magnetic 
flux  is  assumed  to  be  that  indicated  by  the  north-seeking  pole  of  the 
magnetic  needle.  The  magnetic  field 
around  a  current-carrying  wire  is  repre- 
sented, in  direction  and  in  intensity,  in 

Fig.  15- 

4.  The  solenoid. — If  a  current-carrying  FI(J    i6 
conductor  is  wound  into  a  helix,  the  lines 

of  magnetic  induction  are  no  longer  concentric  circles  but  pass 
through  the  helix  as  indicated  in  Fig.  16.     If  the  helix  is  long  in 

*  When  the  conductor  is  isolated  and  uninfluenced  by  other  magnetic  fields. 

29 


30  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

comparison  to  its  diameter,  the  magnetic  flux  is  uniformly  dis- 
tributed over  its  cross-sectional  area  except  near  the  ends  where 
the  lines  begin  to  diverge. 

5.  The  electromagnet.  —  The  magnetic  effect  of  a  current-carry- 
ing wire  is  concentrated  by  winding  it  into  a  helix,  and  greatly  in- 
creased if  the  helix  is  wound  on  an  iron  core.    A  current-carrying  coil 
wound  on  an  iron  core  is  an  electromagnet,  and  is  used  for  the  pro- 
duction of  the  intense  magnetic  fields  required  for  modern  electrical 
apparatus.    When  once  magnetized,  iron  retains  a  part  of  the  mag- 
netic effect  of  the  current-carrying  coil.    This  permanent  effect  is 
termed  residual  magnetism,  and  varies  with  the  quality  of  the 
iron. 

The  polarity  of  an  electromagnet  is  easily  determined  from  the 
direction  of  the  current  in  its  exciting  coil.  If  the  direction  of 
current  around  the  coil  is  that  of  the  rotating  motion  of  a  right- 
handed  screw,  the  direction  of  magnetic  flux  is  that  of  the  longi- 
tudinal movement  of  the  screw. 

6.  Magnetomotive  force.  —  Magnetomotive  force  is  that  prop- 
erty by  virtue  of  which  a  magnetic  flux  is  established  or  maintained. 

7.  Reluctance.  —  The  opposition  offered  to  the  establishment 
or  to  the  maintenance  of  a  magnetic  flux  is  termed  the  reluctance 
of  the  magnetic  circuit.     The  reluctance  of  a  magnetic  circuit  is 
directly  proportional  to  its  length  and  inversely  proportional  to 
its  cross-sectional  area. 

8.  Permeability.  —  Some  materials,  notably  iron  and  many  of 
its  alloys,  offer  less  opposition  to  the  establishment  or  the  main- 
tenance of  a  magnetic  flux  than  do  others.    The  ratio  of  the  flux 
established  or  maintained  in  a  given  length  and  cross  section  of  a 
material  by  unit  magnetomotive  force,  to  the  flux  established  or 
maintained  by  unit  magnetomotive -force  in  the  same  length  and 
cross  section  of  vacuum  is  termed  the  permeability  of  the  material. 

9.  Magnetic  units.  —  The  magnetic  units  are: 

(a)  Unit  pole.  —  Unit  magnet  pole  is  one  which  repels  with  a 
force  of  one  dyne  a  similar  and  equal  magnet  pole  placed  at  a  dis- 
tance of  one  centimeter  hi  air.     (Symbol  m.) 

(b)  Field  intensity.  —  A  magnetic  field  has  unit  intensity  (one 
line  per  square  centimeter  of  cross-sectional  area)  when  it  reacts 
on  a  unit  pole,  placed  in  the  field,  with  a  ^force  of  one  dyne. 
(Symbol  #.) 


MAGNETISM  AND  MAGNETIC  INDUCTION  31 

(c)  Flux  density.  — •  By  flux  density  is  meant  the  number  of  lines 
of  magnetic  force  or  induction  per  unit  of  cross-sectional  area  of  a 
magnetized  material.     (Symbol  B.) 

Note.  —  When  a  magnetic  flux  is  propagated  in  air,  or  other  non- 
magnetic material,  the  flux  density  is  equal  to  the  field  intensity. 

(d)  Magnetic  flux.  —  The  total  flux  in  a  magnetic  circuit  is  equal 
to  the  product  of  the  average  flux  density  and  the  cross-sectional 
area  of  the  circuit.    The  unit  is  the  maxwell.     (Symbol  <£.) 

(e)  Reluctance.  —  Unit  reluctance  (the  oersted)  is  that  opposition 
offered  by  a  cubic  centimeter  of  vacuum,*  each  face  of  which  is  one 
centimeter  square,  to  the  passage  of  a  magnetic  flux  between  its 
parallel  faces.     (Symbol  91.) 

(/)  Magnetomotive  force.  —  The  magnetomotive  force  of  a  circuit 
is  equal  to  the  work  in  ergs  done  when  a  unit  magnet  pole  is 
moved  around  the  circuit  against  the  force,  and  is  measured  in 
gilberts.  (Symbol  cF.)  Unit  difference  of  potential  exists  between 
two  points  when  one  erg  is  required  to  transfer  a  unit  pole  from 
one  point  to  the  other. 

(g)  Magnetizing  force.  —  Magnetizing  force  is  the  ratio  of  the 
magnetomotive  force  of  the  circuit  to  the  length  of  the  circuit. 
Magnetizing  force  is  numerically  equal  to  field  intensity,  and  is 
indicated  by  the  same  symbol. 

(ti)  Permeability.  —  The  permeability  of  a  material  is  the  ratio 
of  the  flux  density  and  the  field  intensity,  or  the  magnetizing  force. 
(Symbol  /*.) 

10.  Law  of  the  magnetic  circuit.  —  The  flux  in  any  magnetic 
circuit  is  directly  proportional  to  the  magnetomotive  force  and  in- 
versely proportional  to  the  reluctance  of  the  circuit. 

;  :?  *-!••   ••  <•> 

when     0  =  the  magnetic  flux  in  maxwells, 

eF  =  the  magnetomotive  force  in  gilberts, 
SH  =  the  reluctance  in  oersteds. 

11.  Faraday's  Law.  —  An  electromotive  force  is  induced  in  any 
closed  circuit  when  the  magnetic  flux  linking  with  the  circuit  changes  in 
value,  and  the  magnitude  of  the  electromotive  force  induced  is  propor- 

*  Air  has  practically  the  same  specific  reluctance  as  vacuum. 


32  ESSENTIALS  OF  ELECTRICAL   ENGINEERING 

tional  to  the  rate  at  which  the  magnetic  flux  linking  with  the  circuit 
changes. 

12.  Lenz*  Law.  —  The  direction  of  an  induced  current  is  such 
that  its  reaction  opposes  any  change  in  the  value  of  the  magnetic  flux 
linking  with  the  closed  circuit,  in  which  the  induced  current  flows. 

Lenz'  Law  is  simply  a  special  application  of  the  general  principle 
of  mechanics  that  action  and  reaction  are  equal  and  opposite. 

13.  Induced  currents.  —  From  Faraday's  Law  it  is  evident  that 
an  electric  conductor  forming  part  of  a  closed  circuit  and  lying 
partly  or  wholly  within  a  magnetic  field  has  an  electromotive  force 
induced  in  it  when:  (a)  the  relative  position  of  the  conductor  and 
the  magnetic  field  changes,  (b)  the  intensity  of  the  magnetic  field 
varies. 

(a)  Induction  by  motion.  —  The  following  propositions  are  self- 
evident  or  are  easily  demonstrated  by  experiment: 

(1)  The  relative  position  of  the  conductor  and  the  magnetic  field 
is  changed  by  the  movement  of  either  the  conductor  or  the  field. 

(2)  The  induction  of  an  electromotive  force  occurs  only  during 
the  period  of  motion. 

(3)  The  induced  electromotive  force  is  proportional  to  the  number 
of  lines  of  magnetic  induction  across  which  the  conductor  cuts  per 
unit  of  time. 

(4)  The  direction  of  the  induced  electromotive  force  is  reversed 
when  the  direction  of  motion  is  reversed.  - 

Let  Fig.  17  represent  a  cross  section  of  a  magnetic  field,  the  flux 
in  which  flows  toward  the  observer,  and  a  copper  conductor  A  which 
may  be  moved  to  the  right  or  to  the  left,  the 
circuit  being  completed  through  the  rails  B, 
C  and  D.  When  the  conductor  A  is  moved  to 
the  right  an  electromotive  force  is  induced, 

the  magnitude  of  which  is  proportional  to 

FlG  I7  the  rate  of  motion,  and  the  direction  of  the 

current  in  the  circuit  is  that  indicated  by 

the  arrows.  If  the  wire  is  moved  to  the  left  at  the  same  speed, 
the  same  value  of  current  flows  in  the  wire  but  its  direction  of  flow 
is  reversed. 

In  the  above  case  the  induced  electromotive  force  is  proportional 
to  the  speed  of  the  conductor,  the  length  of  the  conductor  and  the 
density  of  the  magnetic  field  being  constant.  In  any  construction, 


MAGNETISM  AND   MAGNETIC  INDUCTION  33 

the  number  of  lines  of  magnetic  induction  cut  per  unit  of  time  de- 
pends on: 

(Y)  The  length  of  conductor  lying  within  the  magnetic  field. 

(2')  The  rate  at  which  the  conductor  moves  across  the  magnetic 
field. 

(3')  The  density  of  the  magnetic  field  (the  number  of  lines  of 
magnetic  induction  per  unit  of  cross-sectional  area). 

The  length  of  the  conductor,  the  speed  at  which  the  conductor 
or  the  magnetic  field  moves,  and  the  flux  density  may  be  expressed 
in  any  desired  units,  provided  the  length  of  the  conductor  and  its 
speed  are  in  the  same  unit,  and  the  unit  of  area  is  the  square  of  this 
linear  unit.  The  unit  on  which  the  electromotive  force  of  a  circuit 
is  based  is  that  difference  of  potential  found  to  exist  when  a  con- 
ductor one  centimeter  long  moves*  at  a  uniform  speed  of  one  centi- 
meter per  second  across  a  uniform  magnetic  field  having  a  flux 
density  of  one  maxwell  (one  line  of  magnetic  force  per  square  centi- 
meter). This  unit  of  electromotive  force  is  inconveniently  small 
and  the  commercial  unit  (the  volt)  is  taken  as  100,000,000  times 
the  fundamental  unit  (the  abvolt).  Then 

e  =  I  X  V  X  B  X  io~8,  (2) 

when    e  =  the  electromotive  force  in  volts  induced  in  the  conductor, 
/  =  the  length  of  the  conductor, 
V  =  the  relative  speed  (per  second)  of  the  conductor  and 

the  magnetic  field, 

B  —  the  density  of  the  magnetic  field  (maxwells  per  unit  of 
cross-sectional  area),. 

i.e.,  an  electromotive  force  of  one  volt  is  induced  in  a  closed  circuit 
when  the  magnetic  flux  linking  with  the  circuit  changes  at  the  rate 
of  100,000,000  maxwells  per  second. 

As  pointed  out  above,  a  reversal  of  the  direction  of  motion  causes 
a  reversal  of  the  direction  of  current  flow.  There  is,  then,  a  definite 
relation  between  the  direction  of  motion,  the  direction  of  induced 
electromotive  force  and  the  direction  of  magnetic  flux.  By  refer- 

*  The  direction  of  motion  and  the  axis  of  the  conductor  are  here  assumed  to  be  at 
right  angles  to  each  other  and  to  the  direction  of  the  lines  of  magnetic  force.  When 
this  condition  does  not  exist,  the  effective  length  of  the  conductor  is  equal  to  /  times 
the  sine  of  the  acute  angle  between  the  axis  of  the  conductor  and  that  of  the  magnetic 
field,  and  the  effective  velocity  is  equal  to  V  times  the  sine  of  the  acute  angle  between 
the  direction  of  motion  and  the  axis  of  the  magnetic  field. 


34 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


t&x: 


FIG.  18. 


ence  to  Fig.  17  it  will  be  observed  that  any  one  of  these  three 
quantities  is  at  right  angles  to  each  of  the  other  two.  A  simple 
rule  for  determining  the  relations  of  these  quantities  was  deduced 
by  Dr.  J.  A.  Fleming  and  is  known  as  Fleming's  finger  rule.  If 
the  thumb,  index  and  middle  fingers  of  the  right 
hand  are  placed  as  indicated  in  Fig.  18,  the  thumb 
indicates  the  direction  of  motion  across  the  magnetic 
field,  the  index  finger  the  direction  of  flux,  and  the 
middle  ringer  the  direction  of  induced  electromotive 
force. 

If  a  ring  or  closed  loop  is  rotated  in  a  magnetic 
field,  one  side  of  the  loop  cuts  the  flux  in  one 
direction  and  the  opposite  side  cuts  it  in  the  other  direction. 
Hence,  a  current  flows  around  the  loop  but  its  direction  is  peri- 
odically reversed,  and  the  induced  electromotive  force  passes 
through  successive  values  from  zero  to  maximum  and  then  from 
maximum  to  zero. 

Let  Fig.  19  represent  a  loop  of  wire  and  a  uniform  magnetic 
field,  the  loop  being  rotated  at  a  constant  angular  velocity  in  the 
direction  indicated  by  the  arrow.  Starting  from  the  position 
shown,  the  electromotive  force  increases  to  maximum,  decreases 
to  zero,  increases  to  maximum  in  the  opposite  direction,  and  again 
decreases  to  zero  for  each 
revolution  of  the  loop.  The 
flux  passing  through  (link- 
ing with)  the  loop  decreases 
from  maximum  to  zero,  or 
increases  from  zero  to  max- 
imum, in  one  quarter  of  a 
revolution,  and  the  electro- 
motive force  induced  in  the 
loop  is  proportional,  at  any 
instant,  to  the  sine  of  the 
angle  through  which  the 
loop  has  rotated,  i.e.,  the  electromotive  force  is  harmonic  as  indicated 
in  the  rectangular  representation  where  the  abscissas  are  angular 
distances  (degrees)  passed  through  by  the  rotating  loop,  and  the 
ordinates  are  the  instantaneous  values  of  electromotive  force  in- 
duced in  the  loop.  Since  the  electromotive  force  induced  in  one 


FlG* 


MAGNETISM  AND  MAGNETIC  INDUCTION  35 

side  of  the  loop  tends  to  produce  a  current  in  the  same  direction 
around  the  loop  as  the  electromotive  force  induced  in  the  other  side 
of  the  loop,  the  electromotive  force  of  the  circuit  is  twice  as  great 
as  if  only  one  side  of  the  loop  were  in  the  magnetic  field. 

(b)  Induction  by  varying  the  flux  density.  —  If  two  coils  of  in- 
sulated wire  are  wound  on  an  iron  core  as  indicated  in  Fig.  20,  and 
coil  A  is  supplied  with  a  current  of  con- 
stant value  (continuous  current)  no  cur- 
rent will  flow  in  coil  B.  If  the  current 
in  coil  A  is  made  variable,  the  following 
effects  may  be  noted: 

(1)  A  current  flows  in  coil  B  when  the 
current  in  coil  A  either  increases  or  decreases. 

(2)  A  current  flows  in  coil  B  only  when  the  value  of  the  current 
in  coil  A  is  changing. 

(3)  The  electromotive  force  induced  between  the  terminals  of 
coil  B  is  proportional  to  the  rate  at  which  the  flux  in  the  iron  ring 
changes.    Within  certain  limits,  the  change  in  the  flux  is  approxi- 
mately proportional    to    the   change   in   the   current  flowing   in 
coil  A. 

(4)  The  direction  of  current  in  coil  B  is  reversed  by  reversing  the 
direction  of  current  in  coil  A. 

(5)  The  direction  of  current  in  coil  B  is  always  such  that  it 
opposes  the  change  in  flux  which  produced  it. 

From  the  above  considerations  it  is  evident  that  electrical  energy 
may  be  transferred  from  one  circuit  to  another,  provided  the  cur- 
rent in  the  supply  circuit  is  of  varying  value,  i.e.,  either  pulsating  or 
alternating. 

14.  Reaction  between  a  magnetic  field  and  a  current-carrying 
conductor.  —  Let 

/  =  the  force  required  to  move  a  conductor  across  a  uniform 
magnetic  field  at  a  velocity  of  V  centimeters  per  second, 
B  =  the  density  of  the  magnetic  field  in  maxwells, 
/  =  the  length  of  the  conductor  in  centimeters  lying  in  the 
magnetic  field  and  perpendicular  to  both  the  direction 
of  motion  and  the  direction  of  the  magnetic  flux, 
i  =  the  current  in  amperes  flowing  in  the  conductor, 
e  =  the  electromotive  force  in  volts  induced  in  the  conductor 
(=  eX  io8  c.g.s.  units), 


36  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

ds  =  the  distance  through  which  the  conductor  moves, 

dt  =  the  time  required  for  the  conductor  to  move  through  the 

distance  ds. 

Then  /  (ds)  =  ei  (dt)  joules  (3) 

=  ei  (dt)  X  io7  ergs  (4) 

IBVi  (dt) 
-^ergs  (5) 

'ds      IBVi  ,,. 

and  ;         '*-^'  <6> 

But  |=F.  (7) 

Therefore,  /  =  —  dynes  (8) 

io 

IBi  ,  ^ 


22    B  f      N 

=  -  -pounds.  (io) 

981  Xio4^ 

From  the  above  considerations  it  is  evident  that: 

(a)  A  current-carrying  conductor  lying  within  a  magnetic  field 
tends  to  move  across  the  field,  the  direction  of  motion  being  at 
right  angles  to  the  axis  of  the  conductor  and  to  the  direction  of 
the  flux. 

(b)  The  direction  in  which  the  conductor  tends  to  move  is  re- 
versed by  reversing  either  the  direction  of  the  magnetic  flux  or  the 
direction  of  the  current  flowing  in  the  conductor. 

(c)  The  force  tending  to  move  the  conductor  across  the  magnetic 
field  is  proportional  to  the  product  of  the  current  flowing  in  the 

conductor,  the  density  of  the  magnetic  field,  and  the 
length  of  conductor  lying  in  the  magnetic  field. 

The  relations  between  the  direction  of  the  mag- 
netic  flux,  the  direction  of  the  current  and  the 
direction  of  the  resultant  force  on  the  current-carry- 
ing conductor  are  shown  in  Fig.  21  by  the  index 
finger,  the  middle  finger  and  the  thumb  of  the  left 
hand. 

15.   Magnetic  flux  and  field  intensity  due  to  unit  pole.  —  From 
the  definitions  of  unit  pole  and  field  intensity  given  above,  it 


MAGNETISM  AND   MAGNETIC  INDUCTION  37 

follows  that  the  field  intensity  due  to  unit  pole  is  unity  at  any 
point  on  the  surface  of  a  sphere  one  centimeter  in  radius,  and  of 
which  the  unit  pole  is  the  center.  But  the  area  of  such  a  sphere 
is  4  TT  square  centimeters.  Therefore,  the  total  flux  leaving  or 
entering  a  unit  pole  equals  4  TT  maxwells,  and  the  field  intensity 
due  to  an  isolated  magnet  pole  is  inversely  proportional  to  the 
square  of  the  distance  x  of  a  surface  from  the  pole. 

*-$  («) 

1  6.  Field    intensity    produced    by    an    electric    current.  —  As 

stated  in  Section  3,  the  space  surrounding  a  current-carrying  con- 
ductor is  a  magnetic  field.  The  intensity  of  the  magnetic  field  at 
the  center  of  a  circular  loop,  and  that  surrounding  a  long  straight 
wire,  are  of  particular  interest. 

(a)  Field  intensity  at  the  center  of  a  loop  of  r  centimeters  radius, 
in  which  flows  a  current  of  i  amperes.  —  From  equation  (n),  the 
intensity  of  the  field  set  up  by  a  magnet  pole  m  placed  at  the 
center  of  the  loop  is 


and  the  force  with  which  the  current-carrying  conductor  and  the 
magnetic  field  react  is,  from  equation  (8), 

,      m      2  irri 


,  f    >, 

dynes.  (14) 


If  H"  is  the  field  intensity  at  the  center  of  the  loop  due  to  the 
current  in  the  loop, 

=  mH" 


10  r 


(16) 


J  TT//  O.2    TTl  /          \ 

and  H"  =  --  (17) 

r 

(b)  Field  intensity  x  centimeters  distant  from  the  axis  of  a  long 
straight  wire  in  which  flows  a  current  of  i  amperes.  —  Let  m  magnet 
poles  per  centimeter  length  of  the  wire  be  uniformly  distributed 
along  a  line  parallel  to  and  x  centimeters  distant  from  the  axis  of 


38  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

the  current-carrying  wire.  The  4  TT  lines  which  emanate  from  each 
unit  pole  radiate  like  the  spokes  of  a  wheel,  and  are  in  a  plane 
perpendicular  to  the  axis  of  the  current-carrying  wire.  Therefore, 
the  intensity  of  the  field  at  the  wire  is 


(18) 

2irX 

2m  ,    . 

(19) 

10  x 
and  the  force  exerted  on  each  centimeter  length  of  the  conductor  is 

,      2mi,  ,    , 

/  =  -  —  dynes.  (20) 

IO  OC 

If  H"  is  the  field  intensity  at  the  poles  due  to  the  current-carrying 
wire, 

/  =  mH."  dynes,  (21) 

-r-fft      0.2  mi  /    N 

mH"  =  -  (22) 

1  TT//  O.2  I  ,          \ 

and  H"  =  -  -  -  (23) 

00 

17.  Relation  of  the  gilbert  to  the  ampere-turn.  —  Let  a  current 
of  i  amperes  flow  in  a  coil  of  n  turns.  When  the  coil  is  rotated 
about  a  unit  pole,  or  a  unit  pole  is  moved  around  a  path  threading 
the  coil,  the  work  done  is 


TT7  f    ^ 

W  =  -  (24) 

10 

4  wni  f    \ 

=  -    -ergs.  (25) 

10 

The  magnetomotive  force  due  to  the  current-carrying  coil  is,  by 
definition, 

/  =  m*  (26) 

=  1.257  m  gilberts.  (27) 

The  quantity  ni  is  termed  "  ampere-  turns,"  and  the  ampere- 
turn  is  commonly  used  as  a  unit  of  magnetomotive  force. 

18.  The  C.  G.  S.  unit  of  electric  current.  —  The  c.  G.  s.  unit 
of  current  is  that  current  which,  when  flowing  in  a  wire  in  and 
at  right  angles  to  a  magnetic  field  having  a  density  of  one  maxwell 


MAGNETISM  AND   MAGNETIC  INDUCTION  39 

per  square  centimeter,  causes  a  mechanical  force  of  one  dyne  to 
be  exerted  on  each  centimeter  length  of  the  wire.  From  equation 
(8),  the  c.  G.  s.  unit  is  equal  to  ten  amperes. 

19.  Force  of  magnetic  traction.  —  Lines  of  magnetic  force  act 
like  rubber  threads  under  tension  and  tend  to  shorten,  and  the 
tractive  force  or  pull  between  two  surfaces  in  contact  is  propor- 
tional to  the  square  of  the  flux  density  of  the  magnetic  field  within 
which  the  surfaces  lie. 

Let  the  flux  passing  from  surface  X  (Fig. 
22)  be  divided  into  equal  parts,  and  enter 
poles  m'  and  m"  on  surface  Y.  Pole  m'  lies 
in  the  magnetic  field  of  pole  m"  and  pole  m" 
lies  in  the  magnetic  field  of  pole  m',  and  there  is  a  reaction  between 
each  pole  and  one-half  the  total  flux.  By  definition, 


Therefore,  /'  =  ^  X  f  (29) 

<t>B  , 

(30) 


,    T> 

Similarly,  /"  =  ~-  dynes.  (31) 

Therefore,  /=/'+/"  (32) 

=  ^  dynes.  (33) 

But  in  a  uniform  field  ,    , 

0  =  AB  (34) 

/      AB*J  f    \ 

/  =  —  dynes  (35) 

AB2 


AB2  ,    . 

(37) 


when  A  =  the  area  of  the  surface  in  square  centimeters, 

B  =  the  flux  density  in  maxwells  per  square  centimeter. 

AB2 

/  =  -  -  pounds,  (38) 

72,134,000 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


when  A  =  the  area  of  the  surface  in  square  inches, 
B  =  the  flux  density  in  lines  per  square  inch. 

TABLE   II 
PERMEABILITY  OF  CAST  IRON,  CAST  STEEL  AND  SHEET  STEEL 


Cast  iron 

Cast  steel 

Sheet  steel 

B 

M 

B 

- 

B 

M 

10 

5000 

500 

9,800 

980 

12,500 

I25O 

20 

6600 

330 

13,400 

670 

16,700 

835 

30 

7200 

24O 

14,400 

460 

18,000 

600 

40 

7750 

194 

15,200 

380 

19,000 

480 

50 

8150 

15.700 

3*4 

19,800 

396 

6o 

8500 

142 

16,150 

269 

20,500 

342 

70 

8800 

126 

16,550 

236 

21,  IOO 

301 

80 

9075 

114 

16,925 

211 

21,600 

270 

90 

9325 

IO4 

17,275 

192 

22,000 

244 

IOO 

9600 

96 

17,600 

I76 

22,300 

223 

The  quality  of  iron  varies  greatly,  and  the  above  are  to  be  considered  simply  as 
representative  values. 

20.  Counter-electromotive  force.  —  When  the  reaction  between 
a  current-carrying  conductor  and  a  magnetic  field  causes  the  con- 
ductor to  move  across  the  field,  an  electromotive  force  which  opposes 
the  flow  of  the  current  is  induced  in  the  conductor,  and  the  current 
flowing  in.  the  conductor  is  proportional  to  the  geometrical  dif- 
ference of  the  applied  and  the  induced  (counter)  electromotive 
forces. 

21.  Magnetic  leakage.  —  Unlike  the  electric  current,  a  mag- 
netic flux  cannot  be  confined  to  a  definite  path,  except  under  con- 


(a)  Cb) 

FIG.  23.    Magnetic  Leakage. 

ditions  which  are  not  applicable  to  commercial  electrical  apparatus. 
The  total  flux  in  any  magnetic  circuit  may  be  regarded  as  flowing 
in  two  or  more  parallel  branches,  the  flux  in  each  branch  being 
inversely  proportional  to  the  reluctance  of  that  branch.  In  Fig. 
23  that  part  of  the  flux  which  flows  through  the  air  is  termed  the 
leakage  flux.  The  ratio  of  the  total  flux  set  up  by  a  magneto- 


MAGNETISM   AND   MAGNETIC   INDUCTION 


motive  force  to  the  useful  flux  is  the  leakage  coefficient.  The 
leakage  coefficient  of  a  dynamo  depends  on  its  size  and  design, 
and  varies  from  i.i  to  1.5. 

In  any  magnetic  circuit  containing  iron,  the  leakage  flux  is  in- 
creased by: 

(a)  Increasing  the  total  flux  in  the  circuit.  Since  the  permeability 
of  iron  is  a  function  of  the  flux  density,  the  reluctance  of  the  path 


150 


700       650 


600 


550 


500 


450        400 


550 


.100          150         200         250 
AMPERE-TURNS  PER  INCH 

FIG.  24.    Magnetization  Curves. 


500       550 


through  the  iron  increases  as  the  total  flux  set  up  by  coil  A  in- 
creases, the  reluctance  of  the  path  through  the  air  remains  con- 
stant, and  a  larger  part  of  the  total  flux  is  diverted  through  the 
air.  Fig.  2$&. 

(b)  Increasing  the  length  of  the  air  gap  between  the  metallic  parts 
of  the  magnetic  circuit.  Increasing  the  air  gap  increases  the  reluc- 
tance of  one  branch  of  the  magnetic  circuit  without  affecting  that 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


of  the  other,  and  diverts  a  larger  proportion  of  the  flux  through 
the  leakage  path.     Fig.  230. 

(c)  An  auxiliary  coil  which  sets  up  an  opposing  magnetomotive 
force.  The  counter-magnetomotive  force  set  up  by  coil  B  acts  as 
an  additional  opposition  to  the  flow  of  the  magnetic  flux  through 
that  part  of  the  circuit,  and  increases  the  relative  quantity  of  flux 
diverted  through  the  air.  Fig.  23 c. 

22.  Magnetization  curves.  —  Typical  magnetization  curves  are 
shown  in  Fig.  24  for  cast  iron,  cast  steel  and  annealed  stampings. 

It  will  be  observed  that  for  low  flux  densities  the  magnetization 
curve  is  an  approximately  straight  line,  that  it  bends  sharply  to  the 
right  when  the  magnetizing  force  is  increased  beyond  a  certain  defi- 
nite value,  and  again  becomes  an  approximately  straight  line  if  the 
magnetizing  force  is  sufficiently  increased.  From  these  curves  it  is 
evident  that  the  reluctance  of  a  magnetic  circuit  in  iron  is  not  con- 
stant, but  increases  with  increasing  flux  density. 

23.  The  elementary  dynamo.  —  A  simple  generator  producing 
an  alternating  electromotive  force  may  be  constructed  by  rotating 


an  open  loop  of  copper  wire  or  other  conducting  material  between 
the  poles  of  a  magnet,  and  providing  means  for  connecting  the  termi- 
nals of  the  loop  to  an  outside  circuit.  This  arrangement  is  illus- 
trated in  Fig.  25  where  the  terminals  of  the  loop  are  connected  to 
insulated  copper  rings  mounted  on  the  shaft.  Stationary  conduc- 
tors (brushes)  press  against  these  rings  and  through  them  connec- 
tion is  made  to  an  outside  circuit. 

Unidirectional  currents  are  obtained  in  the  outside  circuit  by  the 
arrangement  shown  in  Fig.  26.     Instead  of  connecting  the  terminals 


MAGNETISM  AND   MAGNETIC  INDUCTION  43 

of  the  loop  to  separate  rings,  connection  is  made  to  a  single  ring 
which  is  divided,  the  segments  being  insulated  from  each  other. 
By  placing  the  brush  mid-way  of  the  pole  face,  it  passes  from  one 
segment  to  the  other  when  the  loop  is  in  the  position  where  it 
is  generating  zero  electromotive  force,  and 
the  current  in  the  outside  circuit  always 
flows  in  the  direction  indicated  by  the 
arrows.  The  magnitude  of  the  electro- 
motive force  varies  with  the  position  of  the  loop,  the  values  during 
one  revolution  of  the  loop  being  plotted  in  Fig.  27.  If,  instead 
of  a  single  loop,  a  number  of  symmetrically  spaced  loops  are  con- 
nected in  series  the  electromotive  force  of  the  system  is  the  sum 
of  the  instantaneous  electromotive  forces 
induced  in  the  individual  loops,  and  the 
potential  difference  between  the  stationary 
brushes  is  approximately  constant.  Fig.  28 
shows  the  electromotive  forces  induced  in 
each  of  three  symmetrically  spaced  loops, 
and  the  resultant  electromotive  force  due 

to  the  combined  action  of  the  three  loops.  The  greater  the 
number  of  loops  in  the  series,  the  more  nearly  the  resultant  approxi- 
mates a  straight  line. 

Structurally  the  electric  generator  and  the  electric  motor  are 
identical,  and  the  same  machine  may  be  used  for  the  production 
of  electrical  energy  or  for  its  conversion  into  mechanical  energy. 

CHAPTER  II  — PROBLEMS 

1.  Find  the  magnetomotive  force  set  up  by  a  winding  of  100  turns  when  a 
current  of  5  amperes  flows  in  the  coil. 

2.  Determine  the  permeability  of  the  iron  around  which  the  coil  in  Problem 
i  is  wound  if  the  length  of  the  coil  is  10  inches,  and  the  flux  density  in  the  iron  is 
10,000  lines  per  square  centimeter. 

3.  The  reluctance  of  a  magnetic  circuit  is  10  oersteds.     Find  the  flux  set  up 
by  a  magnetomotive  force  of  150  gilberts. 

4.  Find  the  flux  density  when  the  force  (pull)  between  an  electromagnet 
and  its  armature  is  60  pounds  per  square  inch. 

5.  A  copper  wire  is  moved  across  a  uniform  magnetic  field  at  the  rate  of  500 
feet  per  minute.     Find  the  electromotive  force,  per  unit  length,  induced  in  the 
wire  if  the  density  of  the  magnetic  field  is  75,000  lines  per  square  inch. 

6.  A  closed  loop  rotates  in  a  uniform  magnetic  field  at  the  rate  of  200  r.p.m. 
and  the  maximum  flux  linking  with  the  loop  is  1,000,000  lines.     Find:  (a)  the 


44  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

effective  value  of  the  electromotive  force  induced  in  the  loop,  (b)  the  current 
flowing  around  the  loop  if  the  resistance  of  the  loop  is  o.oi  ohm.  Neglect  any 
inductance. 

7.  Find  the  force  acting  on  a  current-carrying  wire  10  inches  long,  lying 
wholly  within  a  uniform  magnetic  field  (B  =  45,000  lines  per  square  inch),  and 
carrying  a  continuous  current  of  1000  amperes. 

8.  Find  the  average  force  acting  on  the  wire  in  Problem  7  when  an  alter- 
nating current  of  1000  amperes  (effective)  flows  in  the  wire. 

9.  A  series  magnetic  circuit  is  composed  of:   6  inches  of  cast  iron,  density 
35,000;    8  inches  of  cast  steel,  density  75,000;    j-inch  air,  density  50,000. 
Find  the  ampere-turns  required  in  the  winding.     Use  curves  in  Fig.  24  for  the 
iron  and  the  steel. 

10.  A  magnetic  circuit  10  inches  in  length  is  made  up  of  sheet-iron  punchings 
(laminations),  and  excited  by  1000  turns.     Find  the  exciting  current  when  the 
average  flux  density  is  40,000,  and  90%  of  the  cross-sectional  area  is  iron. 

11.  A  conductor  1 5  inches  long  moves  across  a  magnetic  field  having  a  uni- 
form density  of  48,000  lines  per  square  inch.     Find  the  electromotive  force 
induced  in  the  conductor  if  the  conductor  moves  at  the  rate  of  4500  feet  per 
minute. 

12.  If  the  magnetic  field  in  Problem  n  consists  of  poles  9  inches  long  and 
separated  by  a  distance  of  3  inches,  find  the  average  electromotive  force  induced 
in  the  conductor. 

13.  A  continuous-current  armature  is  18  inches  long  and  30  inches  in  di- 
ameter.    The  electromotive  force  induced  in  each  conductor  as  it  passes  under 
an  interpole  15  inches  long  is  i  volt  when  the  armature  rotates  at  a  speed  of  600 
r.p.m.     Find  the  flux  density  in  the  air  gap  under  the  interpole. 

14.  Show  that  the  field  intensity  in  a  long  solenoid  is 


15-   Show  that  the  ampere- turns  in  the  exciting  coil  of  an  electromagnet 
are 

NI  =  eJL« 

when  B  =  the  maxwells  per  square  centimeter, 

/  =  the  length  of  the  magnetic  circuit  in  centimeters. 

1 6.    Show  that  the  ampere- turns  in  the  exciting  coil  of  an  electromagnet  are 


when  B  =  the  maxwells  per  square  inch, 

/  =  the  length  of  the  magnetic  circuit  in  inches. 

17.  An  electromagnet  lifts  5  tons  (10,000  pounds).  The  mean  length  of  the 
magnetic  circuit  =  60  inches;  cross-sectional  area  =  20  square  inches;  /z  = 
1000.  Find:  (a)  the  total  flux,  (6)  the  flux  density,  (c)  the  ampere-turns  in 


MAGNETISM   AND   MAGNETIC  INDUCTION  45 

the  exciting  coil,  (d)  the  magnetomotive  force  (gilberts),  (e)  the  field  intensity, 
(/)  the  reluctance  of  the  circuit. 

1 8.  A  hollow  wrought-iron  cylinder  (outside  diameter  =12  inches,  inside 
diameter  =  6  inches)  6  inches  long  forms  part  of  a  magnetic  circuit,  the  flux 
in  which  is  parallel  to  the  axis  of  the  cylinder.     Average  density  =  61,000 
lines  per  square  inch,     /z  =20.     Find:    (a)   ampere-turns  required,   (b)   flux 
density  in  the  iron,  (c]  field  intensity,  (d)  reluctance  of  the  cylinder. 

19.  A  point  lies  on  the  axis  of  a  circular  loop  of  wire  in  which  flows  a  cur- 
rent of  i  amperes.     The  radius  of  the  loop  is  r  centimeters,  and  the  point  is 
x  centimeters  distant  from  the  plane  of  the  loop.     Show  that  the  field  inten- 
sity at  the  point  is 

TT         0.2  -n-r-i 
ti  =   - — • 

20.  The  currents  in  two  long  parallel  wires  are  equal  but  flow  in  opposite 
directions.     Show  that  the  field  intensity  at  any  point  on  a  line  joining  the 
centers  of  the  wires  is 

Tr      o.2i  .     o.2i 
a  = h 


D-x 

when  D  =  the  distance  in  centimeters  between  the  axes  of  the  wires, 

x  =  the  distance  in  centimeters  of  the  point  from  the  axis  of  one  wire. 

21.  A  long  solenoid  has  n  turns  per  centimeter  of  its  total  length.  Show 
that  the  field  intensity  at  any  point  inside  the  solenoid,  except  near  the  ends, 
is 

H  =  0.4  irni 

when  i  amperes  flow  in  the  windings. 


CHAPTER  III 
PRACTICAL   CONSTRUCTION   OF   THE   DYNAMO 

i.   Parts.  —  The  principal  parts  of  a  commercial  dynamo  are: 
(a)  frame,  (b)  field  poles,  (c)  field  windings,  (d)  armature  core, 

(e)   armature  winding,    (/)   yoke,    (g)   commutator,    (h)   collector 

rings,  (i)  brushes,  (j)  brush  holders. 


FIG.  29.    Typical  Continuous  Current 
Dynamo.     General  Electric  Co. 


FIG.  30.     Alternating-current  Dy- 
namo.    General  Electric  Co. 


(a)  Frame.  —  The  frame  of  a  dynamo  is  the  supporting  structure, 
and  includes  the  base  and  the  supports  for  the  bearings  in  which  the 
armature  shaft  rests,  as  well  as  the  yoke.     It  is  usually  made  of  cast 
iron  or  cast  steel. 

(b)  Field  poles.  —  The  field  core  is  a  body  of  iron  around  which 
the  field  winding  is  placed.     Its  function  is  to  reduce  the  reluctance 
of  the  magnetic  circuit  (increase  the  flux  for  a  given  field  winding 
and  current),  and  to  supply  a  mechanical  support  for  the  field  wind- 
ing.    It  may  be  made  of  cast  iron,  of  cast  steel,  or  be  built  up  of 
stampings  bolted  or  riveted  together.     Fig.  32  shows  a  typical  form 

46 


PRACTICAL   CONSTRUCTION  OF  THE  DYNAMO 


47 


of  field  pole.    The  part  next  the  armature  is  usually  spread  out  to 

give  a  larger  cross-sectional  area  as  it  is  desirable  to  have  a  smaller 

flux  density  (flux  per  unit  of 

cross-sectional  area)  in  the  air 

gap  than  in  the  body  of  the 

pole.    This   enlarged  part  is 

termed  the  pole  shoe,  and  is 

sometimes  made  as  a  separate 

piece  and  bolted  to  the  core. 
The  field  poles  and  the  yoke 

of  small  machines  are  some- 
times cast  in  a  single  piece. 

When  made  separately,  they 

are  bolted  together,  the  joint 

being  made  as  close  as  possible 

in  order  to  reduce  the  reluc- 
tance of  the  magnetic  circuit. 

(c)  Field  windings.  —  The  field  winding  of  a  dynamo  is  that  part 

which  produces  the  magnetic  field  or  flux  across  which  the  electrical 

conductors  move,  or  which  moves  relatively  to  the  conductors. 
It  consists  (Fig.  33)  of  a  coil  of  insulated  cop- 
per wire,  the  exciting  current  being  supplied 
by  the  dynamo  itself  (self-excited)  or  from  some 
outside  source  (separately  excited). 

(d)  Armature  core. — The  armature   core   is 

FIG.  32.  Typical  Field  that  part  of  the  dynamo  over  or  around  which 
Core  for  Continuous  the  armature  conductors  are  placed.     It  serves 


FIG.  31.     Frame  and  Field  Structure  of  Tri- 
umph Continuous  Current  Dynamo. 


Current    Dynamos. 
Western  Electric  Co. 


as  a  mechanical  support  for   these  conductors 
and   as   a  path  through  which   the   magnetic 

circuit  is  completed.     Armature  cores  are  of  two  classes:  (i)  ring, 

(2)  drum. 

(1)  Ring  cores.  —  The  armature  cores  of  early  dynamos  were  iron 
rings  through  which  the  conductors  were  threaded,  as  shown  in  Fig. 
34a.     Because  of  mechanical  and  electrical  deficiencies,  the  ring 
construction  is  seldom  used  in  present  day  machines. 

(2)  Drum  cores.  —  The  drum  core  consists  of  an  iron  cylinder 
(Fig.  34b)  on  the  surface  of  which  the  armature  conductors  are 
placed. 

Armature  cores  are  built  up  of  thin  stampings  (laminations)  of 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


soft  iron  or  steel.  The  purpose  of  this  construction  is  to  reduce 
the  eddy  currents  which  flow  in  the  body  of  the  core  and  cause  it  to 
heat.  Laminations  vary  in  thickness  from  o.oi  inch  to  0.04  inch. 

Early  dynamos  were 
built  with  smooth  cores, 
the  armature  conductors 
on  their  surface  being  held 
in  place  by  binding  wires. 
In  later  types,  the  arma- 
ture conductors  are  placed 
in  slots  as  shown  in  Fig.  37, 
and  retained  by  means  of 
wooden  or  fiber  wedges. 
Different  slot  forms  are 
used  by  different  manufacturers,  from  an  entirely  closed  slot, 
through  which  the  armature  conductors  are  threaded,  to  the 
rectangular  (open)  slot. 


FIG.  33.    Typical  Field  Coils  (Westinghouse). 


FIG.  34a.    Ring  Armature. 


FIG.  34b.     Drum  Armature. 


(a)  Core.  (b)  Lamination. 

FIG.  35.    Armature  Core  and  Lamination  for  Crocker- Wheeler  Continuous 
Current  Dynamo. 

(e)  Armature   windings. — The   armature   winding   consists   of 
electrical  conductors  which  move  relatively  to  the  magnetic  field. 


PRACTICAL   CONSTRUCTION   OF   THE    DYNAMO 


49 


Complete  Armature  for  Crocker- 
Wheeler  Continuous  Current  Dynamo. 


Retaining  Hedqe 


For  machines  of  the  usual  commercial  voltages,  the  winding  con- 
sists of  several  hundred  insulated  copper  wires  or  bars,  divided 
into  groups,  the  wires  of   each, 
group  being  connected  in  series 
and  the  groups  in  parallel.    The 
number    of    parallel    groups    is 
never  greater  than  the  number 
of  poles  on  the  dynamo  (simplex 
windings),    and    there   may   be 
only  two  parallel   groups  on  a 
continuous-current  armature,  re-  FKJ 
gardless  of  the  number  of  poles. 
The  armature  conductors  of  an 

alternator  are  usually  all  connected  in  series  (single-phase  alter- 
nator).    A  typical  armature  coil  having  one  turn  is  shown  in 

Pig-  38- 

Two  or  more  coil  sides  are  usually  placed  in 
one  slot,  one  side  of  each  coil  being  placed  in 
the  bottom  of  a  slot  and  the  other  side  in  the 
top  of  a  slot  under  an  adjacent  pole.  This 

FIG.  37.    Section  of  arrangement  makes  all  the  coils  on  an  armature 

Slotted    Armature     ...  .      .  ,.        -    .  ,      , 

Core  similar,  and  gives  the  finished  armature  a  sym- 

metrical appearance. 

Continuous-current  armature  windings.  —  The  principal  types  of 
armature  windings  used  on  continuous-current  dynamos  are:  (i)  lap, 
(2)  wave. 

(i)  Lap  or  parallel  windings. 
-The  lap  or  parallel  winding  is 
used  when  the  armature  conduc- 
tors are  to  be  divided  into  as  many 
groups  as  there  are  poles  on  the 
machine.  Starting  at  any  com- 
mutator segment,  the  armature 
conductor  passes  under  one  pole, 
across  the  rear  end  of  the  core,  and  back  under  an  adjacent  pole 
(which  is  of  an  opposite  polarity),  then  to  the  commutator  segment 
adjacent  to  the  one  from  which  the  coil  started.  The  second  coil 
starts  from  the  commutator  segment  at  which  the  first  ends,  the 
third  where  the  second  ends,  etc.,  until  the  winding  closes  on  itself 


FIG.  38.    Typical  Armature  Coil. 
Triumph  Electric  Co. 


50  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

at  the  commutator  segment  from  which  the  first  coil  started. 
Fig.  39- 

Dynamos  having  lap- wound  armatures  are  provided  with  as  many 
brush  sets  as  there  are  poles. 

If,  for  any  reason,  such  as  the  wearing  of  the  bearings,  the  elec- 
tromotive forces  generated  in  the  different  parallel  branches  of  a 
lap-wound  armature  are  unequal,  heavy  currents  circulate  in  the 
winding.  These  currents  cannot  be  prevented  but  their  effects  are 
minimized  by  connecting,  through  heavy  "  equalizer  "  rings,  such 
points  of  the  winding  as  are  normally  of  the  same  potential. 

(2)  Wave  or  series  windings.  —  In  the  wave  or  series  winding,  the 
armature  conductors  are  connected  into  two  parallel  groups.  Start- 
ing at  any  commutator  bar,  the  conductor  passes  under  one  pole, 
across  the  rear  end  of  the  core,  and  back  under  an  adjacent  pole 


s 

FIG.  39.     Elementary  (Lap)  Armature  FIG.  40.     Elementary  (Wave)  Armature 

Winding.  Winding. 

(of  opposite  polarity)  to  a  commutator  bar  which  is  separated  from 
the  one  from  which  the  coil  started  by  a  distance  slightly  greater  or 

v  1-4.1     i       ^       total  number  of  commutator  bars     Tr 
slightly  less  than  -  —  •   If  a  dynamo 

number  of  pairs  of  poles 

has  four  poles,  the  total  number  of  commutator  bars  required  is 

C  =  2F±I,  (i) 

when  Y  is  the  number  of  commutator  bars  between  the  terminal  con- 
nections of  any  given  armature  coil.  If  Y  =  9  (terminals  of  first  coil 
connected  to  commutator  bars  i  and  10),  the  total  number  of  commu- 
tator bars  must  be  either  17  or  19.  By  reference  to  Fig.  40,  it  will  be 
seen  that  the  winding  closes  on  itself,  as  in  the  lap  winding,  by  mak- 
ing a  final  connection  to  the  commutator  bar  from  which  it  started. 


PRACTICAL  CONSTRUCTION  OF  THE  DYNAMO  51 

Since  the  wave  winding  offers  only  two  paths  for  the  current  flow, 
only  two  brush  sets  are  required  and  their  proper  position  is  shown 
in  Fig.  40.  Additional  brushes  may  be  used,  the  total  brush  equip- 
ment acting  like  two  brushes  having  a  total  contact  area  equal  to 
the  sum  of  the  areas  of  the  individual  brushes.  The  additional 
brushes  do  not  affect  the  voltage  to  any  appreciable  extent. 

The  relative  positions  of  the  brushes  on  a  multi-polar  wave- 
wound  armature  make  it  especially  adapted  for  street  car  and 
other  enclosed  motors  which  are  accessible  from  one  side  only. 

Alternating-current  armature  windings.  —  The  armature  winding 
of  an  alternator  does  not  differ,  essentially,  from  that  used  in  con- 
tinuous-current machines,  but  the  winding  does  not  usually  form  a 
closed  circuit.  An  elementary  | 

winding  for  alternators  is 
shown  in  Fig.  41 .  If  the  arma- 
ture rotates,  the  terminals  are 
connected  to  insulated  copper 
rings  mounted  on  the  shaft; 
if  the  armature  is  stationary, 
the  terminals  are  connected 
to  insulated  blocks  from  which 
connection  is  made  to  the 
switchboard. 

Alternator    armature   wind- 
ings in   common  use   are:    (Y)      FlG   4I      Elementary  Armature  Winding 
chain,  (2')  basket.  for  Alternators. 

(V)  Chain  windings.  —  In  the  chain  winding  one  coil  side  only 
is  placed  in  a  slot.  This  winding  is,  therefore,  especially  adapted 
to  high-voltage  machines,  the  coils  being  easily  insulated.  The 
objection  to  the  chain  winding  (but  this  objection  is  not  serious)  is 
that  the  coils  are  of  different  shapes  and  the  individual  conductors 
of  different  lengths.  Fig.  42  shows  a  simple  chain  winding. 

(2'}  Basket  windings.  —  In  the  basket  winding  two  coil  sides  per 
slot  are  used.  The  coils  are,  therefore,  similar,  their  form  and 
arrangement  being  shown  in  Fig.  43.  Basket  windings  are  largely 
used  for  low-  and  medium-voltage  alternators,  and  for  induction 
motors. 

In  both  chain  and  basket  windings  all  the  conductors  are  con- 
nected in  series  (single-phase  windings)  so  that  the  current  flow  is 


N 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


confined  to  a  single  path,  i.e.,  the  terminals  of  one  coil  are  connected 
to  the  terminals  of  adjacent  coils. 

(/)  Yoke.  — The  yoke  is  that  part  of  the  frame  connecting  the 


FIG.  42.    Chain  Armature  Winding. 
General  Electric  Co. 


FIG.  43.     Basket  Armature  Winding. 
Triumph  Electric  Co. 


pole  pieces,  and  serves  the  double  purpose  of  a  mechanical  support 
for  the  field  poles  and  their  windings,  and  of  a  path  for  the  flux  to 
pass  from  the  south  to  the  north  pole.  Fig.  44  shows  cross  sections 

of    several    common    forms. 

The   ridges  on  c  are  added 

to  give  additional  mechanical 

strength. 


(a)  (b)  (c) 

FIG.  44.     Cross  Sections  of  Typical  Yokes. 


(g)  Commutator.  --  The 
commutator  is  a  cylindrical 
structure  made  up  of  copper  bars  insulated  from  each  other  and  from 
the  supporting  structure.  It  is  fastened  to  the  armature  shaft  with 
which  it  rotates.  Its  purpose  is  to  rectify  (make  unidirectional)  the 
alternating  current  which  flows  in  the  armature  winding  of  a  contin- 
uous-current generator,  or  to  periodically  reverse  the  direction  of  the 
current  flowing  in  the  armature  coils  of  a  continuous-current  motor. 
The  best  insulation  obtainable  is  used  to  insulate  the  bars  from 
each  other  and  from  the  supporting  frame.  Mica  is  generally 
used  for  this  purpose.  A  cross  section  and  a  sectional  elevation 


PRACTICAL   CONSTRUCTION   OF  THE  DYNAMO 


53 


of  a  commutator  are  shown  in  Fig.  45,  and  a  typical  commutator 
in  Fig.  46. 

(ti)  Collector  rings.  —  In  an  alternator  with  rotating  armature, 


FIG.  45.    Commutator  Structure. 


FIG.  46.     Commutator.    Tri- 
umph Electric  Co. 


the  terminals  of  the  armature  winding  are  connected  to  insulated 
copper  rings  mounted  on  the  shaft.  In  an  alternator  with  rotating 
field,  similar  rings  (but  not  always  of  copper)  are  provided  for  the 
terminal  connections  of  the  field  windings.  Through  these  rings 
continuous  current  is  supplied  to  the  field. 
Fig.  47- 

(i)  Brushes.  —  The  brushes  are  those  parts 
of  a  dynamo  which  make  sliding  contact  with 
the  commutator  or  the  rings,  and  through 
which  current  is  taken  from  or  supplied  to 
the  rotating  armature,  or  supplied  to  the 
rotating  field  of  an  alternator  with  stationary 
armature.  Brushes  are  made  of:  (i)  carbon, 
(2)  copper. 

(1)  Carbon  brushes.  —  Carbon  brushes  are 

in  very  extensive  use  on  commutating  machines  at  the  present 
time  as  they  offer  a  material  help  in  the  prevention  or  the  reduc- 
tion of  sparking.  Also,  being  set  radially,  the  armature  may  be 
rotated  in  either  direction  without  danger  of  injury  to  the  brushes. 
Carbon  brushes  are  sometimes  coated  with  a  deposit  of  metallic 
copper  to  give  a  reduced  contact  resistance  between  the  brush 
and  the  line  connection. 

(2)  Copper  brushes.  —  Various  forms  of  copper  brushes  are  still 
used  on  alternating-current  apparatus  where  no  commutation  takes 


FIG.  47.   Collector  Rings. 
General  Electric  Co. 


54  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

place,  and  where  it  is  undesirable  to  have  the  large  contact  resist- 
ance characteristic  of  carbon  brushes.  A  common  form  of  copper 
brush  consists  of  a  number  of  thin  leaves  of  spring  copper  soldered 
together  at  one  end. 

(j)  Brush  holders.  —  A  brush  holder  is  the  arrangement  by 
means  of  which  the  brush  is  supported  and  held  in  contact  with 
the  commutator  or  the  ring.  It  is  usually  connected  to  a  rocker 
by  means  of  which  the  angular  position  of  the  brushes  may  be 

changed.  A  typical  brush 
holder  for  carbon  brushes  is 
shown  in  Fig.  48. 

2.  Classes  of  dynamos. 
—  Dynamos  are  divided 
into  two  general  classes 
according  to  the  nature 
of  the  current  they  pro- 
duce or  use:  (a)  continu- 

FIG.  48.     Carbon  Brush  and  Brush  Holder.       QUS    current     (ft)  alternating 
Triumph  Electric  Co. 

current. 

(a)  Continuous-current  dynamos.  —  Continuous-current  dynamos 
are  sub-divided  into  four  divisions  according  to  the  character  of 
the  field  excitation:  (i)  shunt,  (2)  series,  (3)  compound,  (4)  sepa- 
rately excited. 

(i)  Shunt  dynamos.  —  In  the  shunt  dynamo,  the  field  is  excited 
by  means  of  a  winding  composed  of  a  large  number  of  turns  of  insu- 
lated copper  wire  connected  to  the  terminals  of  the  armature  circuit 
so  that  the  field  winding  and  the  load  circuit  are  in  parallel  when  the 
dynamo  is  operated  as  a  generator,  and  the  field  winding  and  the 
armature  circuit  are  in  parallel  when  the  dynamo  is  operated  as  a 
motor. 

The  field  current  is  only  a  small  percentage  of  the  rated  current 
capacity  of  the  dynamo,  and  is  controlled  by  means  of  an  adjustable 
resistance  or  field  rheostat  connected  in  series  with  the  field  winding. 
Since,  for  a  given  magnetization,  it  is  required  that  the  product  of 
the  current  (amperes)  flowing  in  the  winding  and  the  number  of 
turns  in  the  winding  be  constant,  the  larger  the  number  of  turns  the 
smaller  is  the  required  current. 

Fig.  49  shows  the  schematic  and  the  conventional  wiring  diagrams 
of  a  shunt  dynamo. 


PRACTICAL  CONSTRUCTION  OF  THE  DYNAMO 


55 


-FieJd  Resistance 


(2)  Series  dynamo.  —  The  field  of  a  series  dynamo  is  excited  by 
a  few  turns  of  heavy  wire  through  which  flows  the  entire  armature 
current,  or  a  constant  part  of  this  current.  The  field  excitation, 
instead  of  being  approximately  con- 
stant, as  in  the  shunt  dynamo,  is 
proportional  to  the  current  flowing 
in  the  armature,  i.e.,  to  the  load 
on  the  dynamo.  The  field  flux,  how- 
ever, may  not  increase  or  decrease  in 
proportion  to  the  change  in  the  excita- 


FIG.  49.        Wiring  Diagrams  for 
Shunt  Dynamo. 


tion  because  of  properties  of  the  iron  parts  of  the  magnetic  circuit. 
The  schematic  and  conventional  wiring  diagrams  of  a  series  dy- 
namo are  shown  in  Fig.  50.    The  resistance,  shown  in  the  conven- 
tional diagram  as  shunting  the  field  winding,  is  for  the  purpose  of 

varying  the  field  excitation  for  a  given 
armature  current.  This  resistance  is 
made  of  German  silver  or  other  high- 
resistance  material,  and  determines  the 
characteristic  of  the  dynamo.  It  may 
not  be  changed  while  the  dynamo  is 
in  operation,  as  may  the  resistance  in 
the  field  circuit  of  a  shunt  dynamo,  but  is  permanently  connected 
to  the  field  terminals  and  is  changed  only  when  the  characteristic 
of  the  dynamo  is  to  be  altered. 


<  to>  Conventional 


FIG.  50.    Wiring  Diagrams  for 
Series  Dynamo. 


:  b)    SHORT    SHUNT  (conventional) 


<c>    LONG  SHUNT < 


FIG.  51.    Wiring  Diagrams  for  Compound  Dynamo. 

(3)  Compound  dynamo.  —  The  compound  dynamo  is,  as  the  name 
implies,  a  dynamo  having  both  a  shunt  and  a  series  field  winding. 
When  the  shunt  field  winding  is  connected  to  the  terminals  of  the 
armature  circuit,  it  is  termed  a  " short  shunt"  compound  dynamo; 
when  the  shunt  field  winding  is  connected  so  that  the  voltage  be- 
tween its  terminals  is  the  line  voltage,  it  is  termed  a  "long  shunt" 
compound  dynamo.  Conventional  and  schematic  diagrams  are 
shown  in  Fig.  51. 


56  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

(4)  Separately  excited  dynamo.  —  If  the  field  windings  of  a  genera- 
tor are  supplied  from  some  other  source  than  its  own  armature,  or 
the  field  and  armature  of  a  motor  from  different  sources,  it  is  said 
to  be  " separately"  excited.  Continuous-current  dynamos  are 
seldom  separately  excited  while  alternators  are  always  excited  in 
this  way.  The  field  winding  of  a  separately  excited  dynamo  does 
not  differ  from  that  of  a  shunt  dynamo,  and  any  dynamo  may  be 

separately  excited  by  connect- 
mg  its  neld  winding  to  a  con- 
tinuous-current circuit  of  the 
proper  voltage.     Fig.  52. 
(b>  Condon*,  Q)  The    dternator .  —  The 

FIG.  52-     Wiring  Diagrams  for  Separately       ^      essential    difference    be- 
Excited  Dynamo. 

tween  an  alternator  and  a 

continuous-current  dynamo  is  the  substitution  of  continuous  cop- 
per rings  for  the  commutator,  since  alternating  electromotive 
forces  are  induced  in  a  continuous-current  armature.  These  rings 
serve  as  a  means  of  connecting  the  armature  winding  and  the 
outside  or  load  circuit.  The  alternator  is  not  self-exciting. 

However,  in  most  alternators  of  the  present  day,  the  field  instead 
of  the  armature  is  the  rotating  part,  the  armature  winding  being 
placed  in  slots  on  the  inside  of  a  cylindrical  iron  core,  as  shown  in 
Fig.  43.  In  this  type  of  alternator,  rings  are  provided  for  connecting 
the  rotating  field  winding  to  its  continuous-current  supply. 

The  field  excitation  of  an  alternator  is  often  supplied  by  a  small 
continuous-current  generator,  the  armature  of  which  is  mounted  on 
or  belted  to  the  alternator  shaft.  In  large  installations,  the  fields 
of  all  the  alternators  are  supplied  from  one  or  more  (usually  not 
less  than  two)  continuous-current  units,  each  of  which  is  driven 
by  its  own  prime  mover. 

Attempts  have  been  made  to  provide  alternators  with  series  (com- 
pound) windings  by  rectifying  a  portion  of  the  armature  current. 
The  operation  of  such  alternators  (composite  alternators)  is  not 
entirely  satisfactory  *  and  their  manufacture  has  been  discontinued. 

3.  Speed  and  frequency  of  an  alternator.  —  The  speed,  the  fre- 
quency and  the  number  of  poles  of  an  alternator  have  a  fixed  rela- 

*  The  degree  of  compounding  changes  with  the  power  factor,  and  excessive  spark- 
ing takes  place  unless  the  brushes  are  adjusted  so  as  to  pass  from  one  segment  to 
another  at  the  instant  the  current  is  passing  through  its  zero  value. 


PRACTICAL  CONSTRUCTION  OF  THE   DYNAMO 


57 


tion  to  each  other.  An  alternating-current  cycle  means  the  passage 
of  the  current,  or  the  electromotive  force,  through  all  its  values, 
both  positive  and  negative,  i.e.,  starting  at  zero  the  value  rises  to 
maximum,  decreases  to  zero,  rises  to  maximum  in  the  opposite  direc- 
tion, and  again  decreases  to  zero.  To  complete  one  cycle,  a  con- 
ductor must  pass  across  a  north  and  a  south  pole,  or  through  360 
electrical  degrees.*  Then,  for  any  given  frequency,  the  speed  is 
inversely  proportional  to  the  number  of  field  poles, 


p 

and  for  any  given  speed,  the  frequency  is  directly  proportional  to  the 
number  of  field  poles. 

•  ':      ••          /-^,  (3) 

when     /  =  the  frequency  of  the  alternating  current  or  electromo- 
tive force, 

n  =  the  speed  (revolutions  per  second), 
p  —  the  number  of  field  poles. 

In  America  two  frequencies  have  become  standard  —  twenty-five 
cycles  for  exclusive  power  service,  sixty  cycles  for  exclusive  lighting 
service  or  for  service  which  supplies  both  lamps  and  motors.  In 
Europe  a  frequency  of  fifteen  is 
largely  used.  The  lower  frequen- 
cies, while  preferable  in  many  re- 
spects for  motor  operation,  are  en- 
tirely unsatisfactory  for  lamps. 
If  the  frequency  of  an  alternating 
current  supplied  to  incandescent 
lamps  is  reduced  to  about  forty,  a  distinct  variation  in  the  light 
intensity  (which  tires  the  eye)  becomes  noticeable.  The  same 
trouble  is  experienced  with  arc  lamps. 

4.  The  inductor  alternator.  —  An  alternating  electromotive  force 
may  be  induced  in  a  conductor  without  moving  either  the  con- 
ductor or  the  exciting  coil  of  the  field  magnet.  In  Fig.  53  let  A  be 
a  cylindrical  body  of  laminated  iron  similar  to  the  armature  struc- 
ture of  any  rotating  field  alternator,  except  the  central  part  is  cut 

*  An  electrical  degree  is  the  36oth  part  of  the  angle  subtended,  at  the  axis  of  the 
machine,  by  radial  lines  through  the  centers  of  alternate  field  poles. 


FIG.  53.    Schematic  Diagram  of 
Inductor  Alternator. 


58  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

away  to  receive  a  coil  B  of  insulated  wire,  and  C  is  a  body  of  iron 
having  radial  projections  as  shown. 

When  coil  B  is  supplied  with  continuous-current,  a  magnetic  flux 
is  set  up  as  indicated  by  the  heavy  lines  in  the  cross-sectional  view. 
This  magnetic  flux  is  distributed  in  tufts  over  approximately  one- 
half  the  inner  surface  of  part  A  and  moves  around  the  cylinder 
as  part  C  rotates.  The  conductors  on  the  surface  of  A  are  cut  by 
the  flux  in  the  same  manner  as  if  C  were  a  permanent  magnet. 
The  frequency  of  an  inductor  alternator  is 
twice  that  of  a  rotating  field  alternator  having 
the  same  number  of  poles  and  operating  at  the 
same  speed. 

This  construction,  while  very  simple,  fails  to 
show  either  good  regulation  or  high  efficiency. 
5.  Field  discharge  resistance.  —  When  the 
field  circuit  of  a  dynamo  is  opened  there  is 
induced  in  the  field  windings  an  electromotive 
winding  force,  due  to  the  rapid  decrease  in  the  flux 

I QOQOQOQOOOOOOQQQp  J  '  .  _ 

threading  the  windings.    This  induced  electro- 

FIG.  54.  Field  Discharge  r  ,  ,  ,, 

Resistance  motive  force  may  be  so  large  as  to  rupture  the 

insulation  unless  an  auxiliary  resistance,  in 
which  the  energy  stored  *  in  the  magnetic  field  may  be  dissipated,  is 
provided.  Such  a  resistance  is  known  as  a  " field  discharge"  re- 
sistance, and  is  commonly  used  with  machines  having  a  capacity 
greater  than  100  kw.  Fig.  54  shows  the  essential  arrangement 
and  operation  of  a  discharge  resistance. 

CHAPTER  III  — PROBLEMS 

1.  Find  the  resistance  of  a  6-pole  lap- wound  armature,  the  winding  of  which 
consists  of  3000  feet  of  No.  TO  copper  wire  (=  10,400  circular  mils). 

Note.  —  The  resistance  of  any  armature  winding  is  equal  to  the  resistance  of 
the  total  length  of  conductor  on  the  armature,  divided  by  the  square  of  the 
number  of  parallel  paths  into  which  the  conductors  are  connected. 

2.  The  armature  in  Problem  i  is  to  be  wave  wound,  the  same  total  weight 
of  copper  to  be  used  and  the  armature  to  have  the  same  number  of  series  con- 
ductors between  positive  and  negative  brush  contacts.     Find:  (a)  the  size  and 
the  length  of  the  wire  required,  (6)  the  resistance  of  the  armature  winding. 

3.  Plot  a  development  t  of  a  4-pole,  wave- wound  armature  having  35 
slots,  105  commutator  bars  and  630  conductors. 

*  See  Appendix  B,  Section  6. 

t  Armature  windings  should  be  studied  by  means  of  wooden  models  on  which  the 
different  windings  may  be  placed. 


PRACTICAL  CONSTRUCTION  OF  THE   DYNAMO  59 

4.  Plot  a  development  of  a  4-pole,  lap-wound  armature  having  44  slots, 
88  commutator  bars  and  352  conductors. 

5.  The  frequency  of  an  alternator  is  60.    Find  the  speed  when  the  number 
of  poles  equals:   (a)  2,  (b)  4,  (c)  6,  (d)  8,  (e)  10,  (/)  25,  (g)  40. 

6.  The  frequency  of  an  alternator  is  25.     Find  the  speed  when  the  number 
of  poles  equals:  (a)  2,  (b)  4,  (c)  6,  (d)  8,  (e)  10,  (/)  25,  (g)  40.     • 


CHAPTER  IV 

THE  CONTINUOUS-CURRENT  GENERATOR 

1.  The  fundamental  equation.  —  The  voltage  induced  in  the 
armature  windings  of  a  continuous-current  generator  is  directly 
proportional  to:  (a)  the  flux  </>  entering  or  leaving  the  armature  at 
each  pole,  (b)  the  number  of  poles  p  in  the  field  structure,  (c)  the 
total  number  of  conductors  N  on  the  armature,  (d)  the  speed  n 
(revolutions  per  second)  at  which  the  armature  rotates.     It  is  in- 
versely proportional  to  the  number  of  parallel  circuits  p'  into  which 
the  armature  conductors  are  connected.     The  equation  representing 
the  above  statement  is  known  as  the  fundamental^  equation  of  the 
continuous-current  generator. 

„      <{>Nnp     ..  (  ^ 

E  =  Y  ™»  (l) 

The  constant  (io8)  represents  the  ratio  between  the  volt  and  the 
c.G.s.  unit  (the  abvolt)  of  electromotive  force. 

2.  Voltage  characteristic.  —  The  voltage  characteristic  of  a  gen- 
erator is  a  curve  showing  the  relation  between  the  voltage  of  the 
generator  and  the  load  or  the  armature  current,  and  is:  (a)  ex- 
ternal, (b)  internal. 

(a)  External  characteristic.  —  The  external  characteristic  shows 
the  relation  between  the  terminal  voltage  of  a  generator,  and  the 
current  flowing  in  the  load  circuit.     This  is  the  experimental  curve, 
and  is  the  one  usually  referred  to  when  the  term  "  voltage  character- 
istic" is  used. 

(b)  Internal  characteristic.  —  The  internal  or  total  characteristic 
shows  the  relation  between  the  total  voltage  induced  in  the  armature 
winding,  and  the  armature  current.     When  current  flows  in  the  ar- 
mature circuit,  the  resistance  of  the  circuit  makes  the  terminal 
voltage  of  a  generator  less  than  that  induced  in  the  winding.     The 
total  voltage  is,  then,  the  terminal  voltage  plus  the  voltage  drop 
in  the  armature  circuit  and  that  in  the  series-field  windings,  if 
the  dynamo  is  either  series  or  compound. 

+  RJ,  +  Eb>  (2) 

60 


THE   CONTINUOUS-CURRENT   GENERATOR  6 1 

when  Ea  =  the  total  electromotive  force  induced  in  the  armature 

winding, 

Et  =  the  terminal  electromotive  force, 
I a  =  the  armature  current, 
Is  =  the  current  in  the  series-field  circuit, 
Ra  =  the  resistance  of  the  armature  circuit, 
R8  =  the  resistance  of  the  series-field  circuit, 
Eb  =  brush  contact  drop.* 

The  total  current  flowing  in  the  armature  circuit  is,  evidently, 
the  sum  of  that  in  the  load  circuit  and  that  in  the  shunt  field  circuit. 
It  may  be  measured  directly  or  calculated  by  adding  these  two 
quantities.  The  current  in  the  series  field  circuit  of  a  compound 
generator  is  equal  to  that  in  the  armature  circuit  or  to  that  in  the 
load  circuit,  as  the  generator  is  long  or  short  shunt. 

3.  Voltage  regulation.  —  When  the  speed  of  the  armature,  the 
resistance  of  the  armature  winding,!  and  the  resistance  of  the  field 
circuit  are  constant,  the  voltage  regulation  of  a  generator  is  the 
ratio  of  the  maximum  deviation  of  the  actual  characteristic,  between 
full  load  and  no  load,  from  the  ideal  characteristic,  which  is  always 
a  straight  line,  and  the  full-load  (rated)  voltage. 

4.  Building  up.  —  "  Building  up  "  is  that  process  by  which  the 
field  flux  of  a  self -excited  generator  is  increased  to  its  normal  value. 
When  the  iron  in  the  magnetic  circuit  is  once  magnetized  it  retains 
some  of  its  magnetic  property  (residual  magnetism),  and  when  the 
armature  is  rotated  in  this  weak  magnetic  field  a  small  electro- 
motive force  is  induced  in  the  armature  windings.     This  electro- 
motive force  causes  a  current  to  flow  in  the  field  windings  and,  if 
the  windings  are  properly  connected,  the  flux  is  increased.     The 

*  It  has  been  experimentally  determined  that  the  voltage  drop  between  the  brushes 
and  the  commutator  is  a  function  of  the  current  density  in  the  area  of  contact.  The 
drop  per  pair  of  ordinary  carbon  brushes  is  represented,  with  sufficient  accuracy  for 
general  calculations,  by  the  formula: 

£6  =  0.8  +  0.2!),  (3) 

when  D  =  the  current  density  (amperes  per  square  centimeter)  in  the  contact  area 
between  the  brush  and  the  commutator.  Present  practice  allows  five  to  six  amperes 
per  square  centimeter  of  brush  contact  area  at  rated  load.  For  densities  less  than 
one  ampere  per  square  centimeter,  the  values  obtained  by  equation  (3)  are  too  large. 
A  close  approximation  is  obtained  by  assuming  the  brush-contact  drop  equal  to  the 
current  density. 

f  The  brush-contact  resistance  is  not  constant. 


62  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

increased  flux  causes  a  larger  current  to  flow  in  the  field  windings, 
and  increases  the  flux  still  further. 

A  self-excited  generator  can  not  build  up  if  the  flux  set  up  by  the 
field  windings  opposes  the  residual  magnetism  of  the  iron  in  the 
magnetic  circuit.  If  the  magnetism  due  to  the  field  windings 
opposes  the  residual  magnetism,  the  proper  relations  are  obtained 
by:  (a)  reversing  the  field  connections,  (b)  reversing  the  direction  of 
armature  rotation. 

(a)  Reversing  the  field  connections.  —  If  the  terminal  connections 
of  the  field  circuit  are  reversed,  the  direction  of  the  current  flow 
in  the  field  coils  is  reversed,  and  the  two  fluxes  are  no  longer 
opposed. 

(b)  Reversing  the  direction  of  armature  rotation.  —  If  the  direction 
of  armature  rotation  is  reversed,  the  polarity  of  the  armature  ter- 
minals is  changed,  causing  the  direction  of  the  current  in  the  field 
coils  to  reverse,  and  the  two  fluxes  are  no  longer  opposed. 

5.  Commutation.  —  Commutation  is  the  process  of  rectifying  the 
alternating  currents  which  flow  in  the  armature  conductors.  Un- 
like the  elementary  (single  loop)  generator  described  in  Chapter  2, 
Section  23,  the  conductors  of  commercial  armatures  may  carry 
maximum  current  even  though  no  electromotive  force  is  being  in- 
duced in  the  conductors  themselves.  Also,  since  the  current  in  a 
given  coil  periodically  reverses  in  direction,  the  current  in  the  con- 
ductor must  be  reduced  to  zero  and  a  current  of  like  value  estab- 
lished in  the  opposite  direction  during  the  very  short  time  that  the 
brush  is  in  contact  with  the  two  commutator  segments  to  which 
the  terminals  of  the  armature  coil  are  connected.  The  inductance 
of  the  circuit  tends  to  maintain  any  current  that  may  be  flowing 
in  the  coil  at  the  time  the  brush  short-circuits  the  coil,  and  to  pre- 
vent the  establishment  of  a  current  in  the  opposite  direction. 

The  current  flowing  in  the  coil  may  be  reduced  to  zero  and  a  cur- 
rent in  the  opposite  direction  established  during  the  time  the 
brush  is  in  contact  with  two  commutator  segments  by:  (a)  changing 
the  relative  resistances  of  the  paths  through  which  the  current  may 
flow,  (b)  causing  the  conductors  to  generate  an  electromotive  force 
opposite  to  that  which  established  the  original  current. 

(a)  Resistance  commutation.  —  Since  the  resistance  of  an  armature 
coil,  even  when  composed  of  several  turns,  is  very  low,  a  compara- 
tively small  additional  resistance  reduces  the  current  very  materi- 


THE   CONTINUOUS-CURRENT   GENERATOR  63 

ally.  Inherent  properties  of  the  circuit  itself  are  used  for  this 
purpose. 

The  contact  resistance  between  a  carbon  brush  and  the  commu- 
tator is  many  times  the  resistance  of  the  armature  coil  and  increases 
as  the  area  of  contact  decreases.  Let  Fig.  55  represent  the  rela- 
tions of  the  coils,  the  commutator  segments  and  the  brush  of  a  two- 
pole  generator  just  before  the  brush 
makes  contact  with  segment  4.  One- 
half  of  the  line  current  flows  through 
coils  a,  b  and  c,  unites  with  the  other 
half  flowing  in  coils  d,  e  and  /,  and 
passes  to  the  brush  through  the 
radial  connection  and  segment  3 .  An 
instant  later  the  brush  makes  contact 
with  segment  4,  the  current  in  coils  , 

e  and  /  naturally  seeks  the  shorter  path  through  segment  4,  and 
current  is  diverted  from  coil  d.  As  stated  above,  the  inductance  of 
coil'd,  as  well  as  the  relative  resistances  of  the  two  paths  now  pro- 
vided for  the  passage  of  the  current  to  the  brush,  prevents  an  in- 
stantaneous diversion  of  all  the  current  from  the  coil.  As  the 
armature  continues  to  rotate,  the  area  of  the  brush  in  contact  with 
segment  4  increases,  and  the  area  in  contact  with  segment  3  de- 
creases, changing  the  relative  resistances  of  the  two  paths,  so  that 
current  in  coilJ  has  decreased  to  approximately  zero  when  the  area 
of  the  brush  in  contact  with  segment  4  is  equal  to  the  area  in  con- 
tact with  segment  3. 

As  the  area  of  contact  between  the  brush  and  segment  3  decreases 
still  further,  the  resistance  of  this  path  increases,  that  through  seg- 
ment 4  decreases,  and  a  reversed  current  is  gradually  established  in 
coil  d.  When  contact  between  the  brush  and  segment  3  is  broken, 
the  entire  current  should  be  flowing  through  coil  d}  and  no  sparking 
occur. 

(b)  Voltage  commutation.  —  Since  the  electromotive  force  induced 
in  an  armature  conductor  is  zero  only  at  one  point  (the  neutral),  if 
the  brushes  are  advanced  in  the  direction  of  armature  rotation, 
commutation  is  delayed  until  an  opposing  electromotive  force  is 
induced  in  coil  d.  This  opposing  electromotive  force  tends  to 
reduce  the  current  flowing  in  the  coil,  and  to  establish  a  current 
in  the  opposite  direction.  The  magnitude  of  the  opposing  electro- 


FIG.  56.    Commutating  Pole  Field  Struc- 
ture.     Crocker- Wheeler  Co. 


04  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

motive  force  depends  on  the  angular  advance  of  the  brushes. 
Therefore,  to  produce  sparkless  commutation  the  position  of  the 
brushes  must  be  changed  as  the  load  on  the  generator  increases  or 
decreases. 

To  make  the  commutating  flux  proportional  to  the  current  in  the 
armature  coil,  the  construction  shown  in  Fig.  56  is  used.     The  small 

poles  (commutating  or  inter- 
poles) are  excited  by  means  of  a 
winding  connected  in  series  with 
the  load.  The  flux  is,  therefore, 
always  of  the  proper  value  to  pro- 
duce sparkless  commutation,  and 
no  shifting  of  the  brushes  is  nec- 
essary for  changing  load. 

In  nearly  all  dynamos  of  re- 
cent design,  a  combination  of  (a) 
and  (b)  is  used  to  produce  good 
commutation,  high  -  resistance 
carbon  brushes  being  used  in  connection  with  an  angular  advance 
of  the  brushes  or  with  interpoles.  Interpoles  are  extensively  used 
in  present  day  dynamos,  but  satisfactory  commutation,  from  no  load 
to  25  per  cent  overload,  is  accomplished  without  their  use  and 
without  changing  the  position  of  the  brushes. 

6.  Armature  reaction.  —  When 
current  flows  in  an  armature  wind- 
ing, a  magnetic  flux  is  set  up,  as 
indicated  in  Fig.  57.  The  magneto- 
motive force  of  the  armature  is 
proportional  to  the  current  in  the 
winding,  and  the  direction  of  the 
flux  is  at  right  angles  to  the  polar 
axis.  At  one  tip  of  the  pole,  the 
magnetism  thus  set  up  decreases 

the  flux  due  to  the  field  winding,  and  increases  it  at  the  other 
tip,  its  main  effect  being  to  change  the  symmetrical  distribu- 
tion of  the  flux  in  the  air  gap*  shown  in  Fig.  58,  to  the  unsym- 
metrical  distribution  shown  in  Fig.  59,  and  to  shift  the  neutral  (the 


FIG.  57.     Magnetic  Field  Due  to 
Armature  Current. 


*  The  effect  of  armature   teeth  on   the  distribution  of   the   flux  is  here  neg- 
lected. 


THE   CONTINUOUS-CURRENT   GENERATOR 


position  in  which  zero  electromotive  force  is  induced  in  a  conductor) 
in  the  direction  of  armature  rotation. 

To  produce  good  commutation  the  brushes  must  be  moved  fore- 
ward,*  as  indicated  in  Fig.  60.     This  movement  of  the  brushes  pro- 


L.  of  Pole 


CLafPole 


(a) 


(b.) 


FIG.  58.     Symmetrical  Distribution  of  Flux  in  Air  Gap. 


C.L.ofPole 


FIG.  59.    Unsymmetrical  Distribution  of  Flux  in  Air  Gap  Due  to  Armature  Currents. 

duces  a  corresponding  change  in  the  direction  of  the  flux  set  up  by 
the  armature  winding,  and  the  armature  magnetomotive  force  may 
be  resolved  into  two  components  at  right  angles  to  each  other.  One 
of  these  components  is  proportional  to  the  sine  of  the  angle  of  brush 
advance  and  tends  to  set  up  a  flux 
opposite  in  direction  to  that  produced 
by  the  field  windings,  hence  the  term 
"demagnetizing  action."  The  other 
component  of  armature  magnetomotive 
force  is  proportional  to  the  cosine  of 
the  angle  of  brush  advance,  and  has 
the  distorting  effect  described  above. 

The  effect  of  armature  reaction  is, 
then,  twofold:  (a)  Cross-magnetization  or  distortion  which  is  pro- 
portional to  the  current  flowing  in  the  armature  conductors  and  to 
the  cosine  of  the  angle  of  brush  advance.  Because  the  angle  of 
brush  advance  never  exceeds  a  few  degrees,  for  which  the  cosine  is 

*  The  brushes  of  dynamos  having  commutating  poles  are  set  on  the  geometrical 
neutral  (midway  between  the  pole  tips),  and  the  effects  of  armature  reaction  are  re- 
duced to  a  minimum. 


FIG.  60.    Components  of  Arma- 
ture Reaction. 


66  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

approximately  equal  to  unity,  and  the  reluctance  of  the  air  gap 
between  the  pole  tips  is  high,  the  distortion  in  any  given  generator 
may  be  assumed  to  be  proportional  to  the  armature  current. 
(b)  Demagnetization  which  is  proportional  to  the  armature  current 
and  to  the  sine  of  the  angle  of  brush  advance. 

Let  N  =  the  number  of  armature  conductors  on  any  given  arma- 
ture, 

p  =  the  number  of  poles  on  the  dynamo, 
/  =  the  current  flowing  hi  each  armature  conductor. 

Then  the  armature  ampere-turns  per  pole 

=  —  (  ) 

But  the  conductors  are  uniformly  disturbed  over  the  surface  of  the 
armature,  and  the  magnetomotive  force  set  up  by  the  winding  is, 
therefore,  proportional  to  the  average  cosine  (over  180  degrees). 

^  =  ^7 —  av'  cos  I  "  9°o  (5) 

2  P  J  —  90 

^I.257X0.636Ar/*  ^x 

0.4^7    .,,  f  ^ 

=  — * gilberts.  (7) 

7.  The  shunt  generator.  —  The  terminal  voltage  of  a  shunt  gen- 
erator decreases  as  the  load  (armature  current)  increases,  the  speed 
of  the  armature  and  the  resistance  of  the  field  circuit  remaining 
constant.  This  decrease  in  voltage  is  due  to:  (a)  armature  and 
brush-contact  resistance,  (b)  armature  reaction,  (c)  decreased  field 
current. 

(a)  Armature  and  brush-contact  resistance.  —  According  to  Ohm's 
Law,  the  resistance  drop  in  any  current-carrying  conductor  is  equal 
to  the  product  of  the  current  and  the  resistance  of  the  conductor. 
Assuming  a  constant  temperature,  the  resistance  of  the  armature 
winding  is  constant,  and  the  resistance  drop  is  proportional  to  the 
armature  current.  The  drop  due  to  brush-contact  resistance  is 
calculated  by  means  of  equation  (3).  At  no  load,  when  only  the 
field  current  flows  in  the  armature  circuit,  the  armature  current  is 
negligibly  small,  and  the  terminal  voltage  is  equal  to  the  electromo- 
*  See  Appendix  A,  Section  8. 


THE   CONTINUOUS-CURRENT  GENERATOR 


67 


live  force  induced  in  the  armature.  As  the  current  in  the  armature 
increases,  the  resistance  drop  in  the  armature  circuit  becomes  ap- 
preciable, and  the  terminal  voltage  decreases. 

(b)  Armature  reaction.  —  As  explained   in  Section  6,  armature 
reaction  neutralizes  part  of  the  flux  produced  by  the  field  winding 
at  no  load,  and  changes  the  distribution  of  the  flux  in  the  air  gap. 
The  total  electromotive  force  induced  in  the  winding,  and,  there- 
fore,  the   terminal  voltage,   decreases   as   the   armature   current 
increases. 

(c)  Reduction  of  field  current.  —  When  the  resistance  of  the  field 
circuit  is  constant,  the  current  flowing  in  the  circuit  is  proportional 
to  the  electromotive  force  at  the  terminals  of  the  armature,  and  the 
decreased  voltage  due  to  resistance  in  the  armature  circuit  and  to 


£  20  40  60  80          100 

.PER  CENT  OF  RATED  CURRENT 

FIG.  61.    Shunt  Generator  Characteristics. 

armature  reaction,  causes  a  decrease  in  the  field  current,  and  a 
further  decrease  in  the  terminal  voltage  .of  the  generator. 

The  ideal  characteristic  of  a  shunt  generator  is  a  horizontal 
straight  line  (ab  in  Fig.  61),  while  the  actual  characteristic  rises  as 
the  load  decreases  (cd  in  Fig.  61).  The  percentage  regulation  of  a 
shunt  generator  is,  therefore, 

*  — 


per  cent  regulation 

when  E  =  the  full-load  (rated)  voltage, 
EQ  =  the  no-load  voltage. 


(8) 


The  regulation  of  the  average  shunt  generator  is  too  large  for  the 
satisfactory  operation  of  lamps  when  the  load  varies  greatly,  al- 
though it  may  be  entirely  satisfactory  for  motors.  To  give  satis- 
faction, incandescent  lamps  must  have  approximately  constant 
voltage  applied  between  their  terminals.  Since  the  heating  of  a 
conductor  is  proportional  to  the  square  of  the  current  flowing  in  it, 


68 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


& 


100 


80 


:  60 


40 


20 


the  light  given  by  an  incandescent  lamp  is  approximately  propor- 
tional to  the  square  of  the  applied  voltage.  The  inherent  regulation 
of  a  shunt  generator,  therefore,  cannot  be  depended  on  to  maintain 
the  proper  voltage  as  the  load  fluctuates,  and  manipulation  of  the 
field  rheostat  must  be  resorted  to  when  incandescent  lamps  form  all 
or  part  of  the  load. 

Automatic  adjustment  of  the  field  resistance  is  made  by  the  Tirrill 
and  other  regulators,  which  are  largely  used  for  the  maintenance  of 
constant  voltage  either  at  the  generator  terminals  or  at  some  center 

of  distribution. 
8.  The  series  generator. 
-  The  voltage  characteristic 
of  a  series  generator  is,  prac- 
tically, the  magnetization 
curve  for  a  combined  iron 
and  air  circuit.  Since  the 
load  current,  or  a  constant 
part  of  it,  flows  through  the 
series  -  field  windings,  the 
field  excitation  increases  as 
the  load  increases.  For  low 
excitations,  the  flux  is  very 
nearly  proportional  to  the 
field  current,  but  as  the  ex- 
citation increases,  the  flux 
increases  at  a  constantly 
decreasing  rate  until  the  iron 
becomes  "saturated."  Be- 
yond the  point  of  saturation,  the  permeability  of  iron,  is  little  greater 
than  that  of  air,  and  a  large  increase  in  field  current  produces  only 
a  small  increase  in  flux.  The  voltage  of  a  series  generator  is,  there- 
fore, approximately  proportional  to  the  armature  current  over  a 
considerable  range,  the  characteristic  bending  to  the  right,  as  indi- 
cated in  Fig.  62. 

Armature,  brush-contact  and  field  resistances  make  the  terminal 
voltage  less  than  the  induced  voltage,  while  armature  reaction  re- 
duces the  total  flux  set  up  by  a  given  field  excitation  and  causes 
an  unsymmetrical  distribution  of  the  flux  in  the  air  gap  as  in  other 
types  of  generators. 


20 


40    60    80    100  170  140   160   180  ?00 

PER  CENT  OF  RATED  CURRENT 


FIG.  62.     Series  Generator  Characteristics. 


THE   CONTINUOUS-CURRENT  GENERATOR  69 

Since  the  series  generator  is  used  almost  exclusively  as  a  constant- 
current  generator,  its  voltage  regulation  is  of  little  consequence. 
The  term  "regulation,"  when  applied  to  a  series  generator,  means 
the  ratio  of  the  maximum  deviation  of  the  current,  between  rated 
load  and  short  circuit,  from  the  rated  current  at  full  load,  and  the 
rated  current. 

When  used  as  a  constant-current  generator,  the  series  dynamo  is 
provided  with  an  automatic  regulator  which  causes  the  terminal 
voltage  to  increase  or  decrease  in  proportion  to  the  increase  or  de- 
crease in  the  resistance  of  the  load  circuit,  so  that  the  current  is 
maintained  at  an  approximately  constant  value.  One  of  the 
simplest  of  these  automatic  regulators  is  that  used  on  the  Brush  arc- 
light  generator.  The  field  winding  is  shunted  by  a  carbon  pile,  the 
resistance  of  which  varies  inversely  as  the  pressure  between  the  discs. 
The  pressure  between  the  discs  is  varied  by  means  of  an  electro- 
magnet, the  winding  of  which  is  connected  in  series  with  the  load. 
When  normal  current  flows  in  the  circuit,  the  current  divides,  part 
flowing  in  the  field  windings  and  part  through  the  carbon  pile.  If 
the  current  falls  below  normal,  by  reason  of  an  increase  in  the  resist- 
ance of  the  load  circuit  (an  increase  in  the  number  of  lamps  in  the 
circuit),  the  pressure  on  the  carbon  pile  is  decreased.  The  de- 
creased pull  of  the  magnet  increases  the  resistance  of  the  carbon 
shunt,  and  causes  a  larger  current  to  flow  in  the  field  windings. 
This  increased  field  excitation  causes  a  larger  voltage  to  be  induced 
in  the  armature  windings,  and  the  load  current  rises  to  its  normal 
value. 

Since  no  current  flows  in  the  field  coils  of  a  series  generator  on 
open  circuit,  it  can  " build  up"  only  when  the  load  circuit  is  closed. 

9.  The  compound  generator.  —  The  inherent  tendency  for  the 
terminal  voltage  of  a  shunt  generator  to  decrease  as  the  load  in- 
creases is  counteracted  by  the  addition  of  a  series-field  winding 
so  proportioned  that  the  total  flux  increases  as  the  current  in  the 
armature  winding  increases.  If  the  effect  of  the  series- field  wind- 
ing is  just  sufficient,  at  full  load,  to  compensate  for  the  decrease  in 
voltage  due  to  armature  resistance  and  armature  reaction  (i.e.,  if 
the  no-load  and  the  full-load  voltages  are  equal),  the  generator  is 
flat  compounded;  if  the  effect  of  the  series-field  winding  is  such 
that  the  full-load  voltage  is  greater  than  the  no-load  voltage,  the 
generator  is  over  compounded.  The  characteristic  curves  of  a  flat- 


70  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

compounded  generator  and  of  an  over-compounded  generator  are 
shown  in  Fig.  63. 

When  armature  speed,  field  resistance  and  armature  resistance 
are  constant,  the  compounding  of  a  generator  is  the  ratio  of  the  in- 
crease in  voltage,  between  no  load  and  full  load,  to  the  no-load 
voltage. 


Per  cent  compounding  = 

when    EQ  =  the  no-load  voltage, 
E  =  the  full-load  voltage. 


(E  -  EQ)  100 


(9) 


It  is  impracticable  to  build  a  compound  generator,  the  voltage 
characteristic  of  which  is  a  straight  line.  From  the  definition  given 
in  Section  3,  the  regulation  of  a  compound  generator  is  the  maximum 
deviation  e  (Fig.  63),  between  no  load  and  full  load,  of  the  char- 


20          40          60  ftO          100          120         140 

PER  CENT  OF   RATED  CURRENT 

FIG.  63.     Compound  Generator  Characteristics. 

acteristic  from  the  straight  line  connecting  the  no-load  and  the  full- 
load  voltage  points,  divided  by  the  full-load  voltage.  The  maximum 
deviation  is  measured  perpendicularly  to  the  axis  of  abscissa,  and 
only  for  the  flat-compound  generator  is  it  perpendicular  to  the 
ideal  curve. 


Per  cent  regulation  =  — —  , 


(10) 


when  E  =  the  full-load  voltage. 


A  compound  generator  builds  up  in  the  same  manner  as  a  shunt 
dynamo  since  the  excitation  due  to  the  series-field  winding  is  negli- 
gible at  no  load.  If  the  series  winding  is  improperly  connected,  the 
two  windings  oppose  each  other  magnetically,  and  the  terminal  volt- 
age drops  very  fast  as  the  resistance  of  the  load  circuit  is  decreased. 


THE  CONTINUOUS-CURRENT  GENERATOR  71 

The  degree  of  compounding  of  a  given  generator  may  be  changed, 
without  altering  its  structure,  by:  (a)  changing  the  resistance  of  the 
shunt  around  the  series-field  winding,  (b)  changing  the  speed. 

(a)  Changing  the  resistance  of  the  series-field  shunt.  —  The  effect  of 
changing  the  resistance  shunting  the  series-field  winding  is  to  cause 
a  greater  or  a  less  percentage  of  the  total  load  current  to  flow  in 
the  series-field  coils,  thus  increasing  or  decreasing  the  field  excitation 
for  a  given  armature  current.    This  change  in  field  excitation  causes 
the  full-load  voltage  to  be  increased  or  decreased. 

(b)  Changing  the  speed.  —  If  the  no-load  voltage  of  the  generator 
remains  constant,  the  effect  of  an  increase  in  the  speed  of  its  arma- 
ture is  to  increase  the  full-load  voltage.     The  effect  of  a  decrease 
in  the  speed  of  a  given  generator,  the  no-load  voltage  of  which  is 
constant,  is  to  decrease  the  full-load  voltage. 

The  truth  of  these  statements  is  evident  from  a  study  of  the  oper- 
ating principles  of  the  electric  generator.  The  rated  full-load  volt- 
age of  a  flat  compounded  generator  is  100.  At  no  load,  1,000,000 
magnetic  lines  pass  into  the  armature  from  each  north  pole,  and 
this  flux  is  increased,  at  full  load,  to  1,100,000  lines  by  the  magnetic 
action  of  the  series-field  winding.  The  electromotive  force  induced 
in  the  armature  at  full  load  is,  therefore,  no,  and  10  volts  are  re- 
quired to  compensate  for  armature  resistance  and  armature  re- 
action. 

If  the  speed  of  the  armature  is  increased  20  per  cent  above  its 
rated  value,  the  flux  required  to  produce  the  no-load  voltage  is  re- 
duced to  800,000  lines,  while  the  effect  of  the  series  winding  does 
not  decrease.  Assuming  the  magnetic  effect  of  the  series  winding 
to  be  the  same  at  all  speeds,  the  induced  electromotive  force,  at 
the  higher  speed,  is  112.5  and  the  terminal  voltage,  at  full  load,  is 
102.5,  an  over  compounding  of  2.5  per  cent. 

But  the  compounding  is  greater  than  the  above  because  of  the 
fact  that  as  the  shunt-field  excitation  is  reduced,  the  flux  produced 
by  a  given  series-field  excitation  increases.  The  last  statement 
will  be  made  clear  by  an  examination  of  the  magnetization  curve 
of  a  generator.  Referring  to  Fig.  64,  it  is  found  that  500  ampere 
turns  are  required  to  produce  a  flux  of  1,000,000  lines,  and  that 
850  ampere- turns  are  required  to  produce  1,100,000  lines.  Conse- 
quently, the  shunt-field  winding  must  consist  of  500  ampere-turns 
and  the  series-field  winding  of  350  ampere- turns.  To  produce 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


800,000  lines  requires  only  260  ampere-turns  in  the  shunt-field 
winding,  but  the  series-field  winding  still  produces,  at  full  load,  350 

l.200,000r 


^  1,000.000 


2    800.000 
o 

I 


600.000 


«o    400,000 


Z  00.000 


200       400          600         800         1000        1200        1400 
MAGNETOMOTIVE   FORCE  IN  AMPERE  TURNS. 

FIG."  64.  Characteristic  magnetization  curve  showing  the  large  increase  in  magnet- 
izing force  for  a  small  increase  in  magnetic  induction  after  passing  the  "knee"  of 
the  curve. 

ampere-turns  or  a  total  of  610  ampere- turns  at  full  load.  This  ex- 
citation sets  up  a  flux  of 
1,030,000  lines,  the  elec- 
tromotive force  induced  in 
the  armature  at  full  load 
is  128.5,  and  the  terminal 
voltage  is  118.5,  an  over 
compounding  of  18.5  per 
cent. 

10.  The   Tirrill    regu- 
lator. —  An     elementary 


.•Compenaattfty  Resistance  Shunt 


FIG.  65.    Diagram  of  Connections  for  Tirrill 
Regulator. 


diagram  of  the  Tirrill  regulator,  as  applied  to  a  shunt  or  a  compound 
generator,  is  shown  in  Fig.  65.  By  means  of  electromagnetically 
operated  contacts,  which  open  and  close  as  the  electromotive  force 


THE   CONTINUOUS-CURRENT   GENERATOR  73 

rises  above  or  falls  below  a  certain  value  at  the  generator  terminals, 
or  at  some  center  of  distribution,  the  field  rheostat  is  periodically 
short-circuited. 

The  winding  of  the  main  contact  magnet  is  connected  directly  to 
the  bus  bars,  and  is  so  proportioned  that  it  opens  the  main  con- 
tacts, against  the  action  of  a  spring,  when  the  voltage  rises  above 
the  desired  value.  The  relay  is  differentially  wound,  one  coil  being 
connected  directly  to  the  bus  bars  and  the  other  in  series  with  the 
main  contacts.  The  relay  contacts  are  connected  to  the  terminals 
of  the  field  rheostat.  The  function  of  the  condenser  is  to  prevent 
excessive  sparking  when  the  relay  contacts  open. 

The  regulator  operates  as  follows :  If  the  main  contacts  are  open, 
the  left-hand  coil  of  the  differentially-wound  relay  is  energized 
and  the  relay  contacts  opened.  This  action  increases  the  resistance 
of  the  field  circuit  by  opening  the  short  circuit  over  the  field  rheo- 
stat, the  voltage  drops  below  the  required  value,  and  the  spring 
closes  the  main  contacts.  Closing  the  main  contacts  energizes  the 
right-hand  coil  of  the  relay  and  closes  the  relay  contacts,  thus  raising 
the  voltage  above  the  value  at  which  the  main  contacts  open.  In 
practice,  the  two  sets  of  contacts  are  continually  opening  and  clos- 
ing, and  the  voltage  varies  from  the  desired  value  by  only  a  very 
small  amount. 

When  it  is  desired  to  maintain  constant  voltage  at  some  distant 
center  of  distribution,  a  differential  series  winding  is  added  to  the 
main  control  magnet,  thus  increasing  the  bus-bar  voltage  at  which 
the  main  contacts  are  opened.  The  drop  in  the  transmission  sys- 
tem may  thus  be  compensated  for. 

n.  Parallel  operation  of  generators.  —  It  is  often  desirable  to 
supply  an  electric  system  from  two  or  more  generators  instead  of 
from  one  because:  (a)  of  the  greater  efficiency  of  a  machine  when 
operated  at  or  near  its  rated  load,  (b)  continuity  of  service  can  be 
more  easily  maintained. 

(a)  Efficiency.  —  Because  certain  losses  of  a  generator  are  practi- 
cally constant  and  these  losses  are  roughly  proportional  to  the  size  of 
the  generator,  the  ratio  of  the  total  output  to  the  total  input  (all-day 
efficiency)  is  greater  for  a  generator  running  at  full  load  than  for  one 
operating  at  half  its  rated  capacity,  the  total  output  being  constant. 

(b)  Continuity  of  service.  —  It  is  practically  impossible  to  main- 
tain continuous  service  from  a  single  generator  because  adjustments 


74 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


and  repairs,  which  require  the  machine  to  be  shut  down,  must  be 
made  from  time  to  time.  With  two  or  more  generators,  these 
matters  may  be  attended  to  during  periods  of  light  load  when  all 
the  machines  are  not  required.  It  is  thus  possible  to  give  con- 
tinuous service  and,  at  the  same  time,  keep  the  machinery  in  repair. 
Parallel  operation  of  generators,  then,  means  dividing  the  total 
load  between  two  or  more  generators,  and  the  operation  of  such  a 
number  of  generators  at  a  time  as  will  cause  each  generator  to 
operate  at  its  highest  efficiency.  This  necessitates  connecting  and 
disconnecting  generators  and  the  load  circuit  as  the  load  varies. 

Before  two  continuous-current  generators  are  connected  to  the 
same  load  circuit,  their  terminal  voltages  should  be  approximately 
equal  and  similar  terminals,  either  positive  or  negative,  must  be  con- 
nected together.  If  the  voltages  are  not  approximately  equal,  an 
undesirable  surge  of  current  takes  place  in  the  system  when  the 
switch  is  closed.  When  dissimilar  terminals  of  the  generators  are 
connected  together,  conditions  identical  with  a  short-circuit  exist 
and  an  excessive  current,  which  either  operates  the  protective  de- 
vices or  overheats  the  armatures,  flows  around  the  circuit  formed 
by  the  two  armature  windings. 

Shunt  generators.  —  The  connections  for  the  operation  of  two 
shunt  generators  in  parallel  are  indicated  in  Fig.  66.  If  generator 

A  is  carrying  the  load  and 
it  is  desired  to  divide  the 
load  between  A  and  B, 

I  drive   generator  B   at   its 

rated  speed  and  regulate  its 
'  field  to  give  a  voltage  equal 
to  or  slightly  greater  than 
FIG.  66.  Parallel  Operation  of  Shunt  Generators,  f^at  of  A  When  the  volt- 
age across  the  open  switch  is  zero,  the  switch  may  be  closed,  after 
which  the  field  rheostats  should  be  manipulated  until  the  volt- 
age of  the  system  is  the  rated  or  required  value,  and  the  genera- 
tors divide  the  total  load  in  proportion  to  their  ratings.*  After 
being  properly  regulated,  two  or  more  generators  having  similar 
characteristics  will  automatically  divide  the  load  in  proportion  to 
their  ratings  as  the  total  load  on  the  system  varies. 


*  The  load  on  any  one  of  two  or  more  continuous-current  generators  operating  in 
parallel  is  reduced  by  reducing  either  its  field  excitation  or  its  speed. 


THE   CONTINUOUS-CURRENT  GENERATOR 


75 


In  connecting  two  generators  as  described  above,  a  voltage  across 
the  switch  indicates  that  the  terminal  connections  of  generator  B 
should  be  reversed.  When  any  voltage  is  indicated  across  this 
switch,  it  is  equal  to  approximately  twice  the  voltage  of  each  gen- 
erator, so  that  in  testing,  care  should  be  taken  that  the  voltmeter  is 
not  injured. 

A  shunt  generator,  operating  in  parallel  with  other  generators, 
should  be  disconnected  from  the  load  circuit  by  opening  the  line 
switch  or  circuit-breakers  after  reducing  either  its  speed  or  its  field 
excitation  until  the  current  flowing  in  its  armature  is  a  minimum. 

Compound  generators.  —  The  connections  for  the  parallel  opera- 
tion of  compound  generators  are  shown  in  Fig.  67,  the  only  change 
from  the  shunt  diagram  being  a  third  connection  between  the  two 
generators.  This  connec- 
tion is  called  the  equal- 
izer, and  is  made  between 
the  brush  and  the  series- 
field  winding  of  each  gen- 
erator. 

The  effect  of  the  equal- 
izer is  to  divide  the  load 
properly  between  the  two 
generators.  Consider  the 


Field  Resistance 


Field 


FIG.  67. 


Parallel  Operation  of  Compound 
Generators. 


action  of  two  compound  generators  without  an  equalizer  connection. 
If,  for  any  reason,  the  speed  of  one  generator  increases  slightly,  its 
voltage  increases  and  it  takes  a  greater  part  of  the  total  load.  If 
the  total  load  on  the  system  remains  constant,  the  increased  load  on 
one  generator  causes  a  larger  current  to  flow  in  its  series-field  wind- 
ing and  reduces  the  current  flowing  in  the  series-field  winding  of 
the  other  generator.  This  change  in  field  excitation  disturbs  the 
equilibrium  of  the  generators  and  will,  by  its  cumulative  effect, 
cause  one  generator  to  take  the  entire  load  and  to  operate  the 
other  as  a  motor.  Compound  generators  connected  to  the  same 
circuit  and  operating  without  equalizer  connections  are,  therefore, 
in  unstable  equilibrium. 

With  an  equalizer  connection,  the  same  increase  in  the  speed  of 
generator  A  causes  its  voltage  and  its  current  output  to  increase, 
but  instead  of  all  the  increased  current  flowing  through  the  series- 
field  winding  of  generator  A,  it  divides  at  the  brush,  part  flowing 


76  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

through  the  series-field  winding  of  each  generator  so  that  the  voltage 
of  each  is  equally  increased,  and  their  equilibrium  is  not  disturbed. 

Equalizer  connections  should  be  large  as  they  are  often  required 
to  carry  heavy  currents,  and  an  appreciable  resistance  in  this  con- 
nection between  generators  tends  to  disturb  their  equilibrium,  and 
causes  an  unequal  division  of  the  load.  Fuses  or  other  circuit  in- 
terrupting devices  should  not  be  placed  in  an  equalizer  circuit. 

The  process  of  disconnecting  a  compound  generator,  operating  in 
parallel  with  other  generators,  from  its  load  circuit,  is  similar  to  that 
for  a  shunt  generator. 

12.  Connection  to  load  circuits.  —  The  load  units  supplied  by 
shunt  and  compound  generators  are  connected  in  parallel,  i.e.,  the 
terminals  of  each  unit,  as  a  lamp  or  a  motor,  are  connected  directly 
to  the  mains  leading  from  the  terminals  of  the  generator  or  to 
branches  of  these  mains. 

The  shunt  generator  is  in  common  use  under  the  following  con- 
ditions: 

(a)  Where  the  load  is  practically  constant. 

(b)  Where  the  load  changes  slowly  and  infrequently. 

(c)  Where  the  load  consists  exclusively  of  motors  which  do  not 
require  close  voltage  regulation. 

(d)  Where  an  attendant  is  employed  to  manipulate  the  field  rheo- 
stat and  maintain  constant  voltage. 

(e)  Where  the  voltage  is  controlled  by  a  Tirrill  regulator  or  other 
automatic  device. 

Compound  generators  are  in  common  use  under  these  conditions: 

(af)  Where  an  approximate  compensation  is  to  be  automatically 
made  for  the  internal  drop  of  the  generator. 

(b')  Where  the  drop  in  a  long  transmission  line  or  in  a  long  feeder 
is  to  be  automatically  compensated  for,  i.e.,  where  the  voltage  at 
the  terminals  of  a  load  some  distance  from  the  generator  is  to  be 
automatically  maintained  at  an  approximately  constant  value  as 
the  load  varies. 

(c')  Where  the  terminal  voltage  is  to  increase  with  the  load. 

The  series  generator  is  used  almost  exclusively  for  supplying  cur- 
rent to  series  arc-lamp  systems,  the  generator  being  provided  with  a 
regulator  which  automatically  maintains  the  current  in  the  system 
at  an  approximately  constant  value  as  the  load  changes. 

The  terminals  of  the  load  units  in  a  series  circuit,  instead  of  being 


THE   CONTINUOUS-CURRENT  GENERATOR  .  77 

connected  to  the  mains  running  from  the  generator,  are  connected 
to  each  other,  one  terminal  of  the  first  and  one  terminal  of  the  last 
unit  in  the  series  being  connected  to  the  mains.  Consequently,  the 
same  current  flows  in  each  part  of  the  circuit  and  the  voltage  changes 
as  the  load  changes. 

The  series  generator  for  the  operation  of  arc-lamp  circuits,  has 
been  largely  replaced  by  alternating-current  apparatus. 

CHAPTER  IV  — PROBLEMS 

1.  The  electromotive  force  of  a  shunt  generator  is  118  volts  when  operated 
without  load,  and  the  rated  output  is  n  kw.  at  no  volts.     Calculate:  (a)  the 
voltage  regulation,  (6)  the  resistance  of  the  armature  circuit,  if  one-half  the 
drop  in  terminal  voltage  between  no  load  and  full  load  is  due  to  armature 
resistance. 

2.  The  terminal  voltage  of  a  series  generator  is  450  when  the  armature  cur- 
rent is  10  amperes.     Calculate  the  electromotive  force  induced  in  the  armature 
winding  if  the  resistance  of  the  armature  circuit  is  i  ohm  and  that  of  the  field 
winding  is  1.5  ohms. 

3.  A  shunt  generator  requires,  at  zero  load,  a  field  current  of  2  amperes  to 
induce  the  voltage  at  which  it  is  rated;  at  full  load  (=  100  amperes)  the  field 
current  required  to  give  rated  terminal  voltage  is  2.33  amperes.     The  shunt 
winding  on  each  pole  consists  of  4000  turns.     Find  the  number  of  turns  re- 
quired in  a  series-field  winding  to  make  the  generator  flat  compounded,  57  per 
cent  of  the  load  current  to  flow  in  the  series-field  windings. 

4.  A  4-pole,  i25-volt,  lap-wound  (drum)  armature  has  400  conductors. 
The  flux  per  pole  is  2,500,000.     Find  the  speed  at  which  the  armature  rotates 
to  induce  rated  voltage  in  the  armature  conductors. 

5.  A  i6-pole,  lap-wound  generator  operates  at  80  r.p.m.,  has  a  flux  per  pole 
of  7,500,000,  and  2304  conductors.     Find  the  voltage  induced  in  the  armature. 

.  6.   Find  the  number  of  conductors  required  on  a  wave-wound  armature  to 
be  used  in  the  generator  of  Problem  4,  the  rated  voltage  to  remain  the  same. 

7.  Compare  the  resistance  of  a  6-pole,  lap-wound  armature  with  that  of  a 
6-pole  wave-wound  armature,  the  rated  voltages,  the  flux  per  pole,  the  kw. 
outputs,  and  the  current  densities  in  the  armature  conductors  being  the  same. 

8.  Allowing  500  circular  mils  per  ampere,  find  the  area  of  the  armature 
conductors  in:  (a)  a  6-pole  lap-wound  dynamo,  the  armature  current  of  which 
is  100  amperes,  (b)  a  6-pole  wave-wound  dynamo,  the  armature  current  of 
which  is  100  amperes. 

9.  Find  the  voltage  induced  in  an  8-pole  lap- wound  armature  when: 

r.p.m.  =  300,  <f>  =  2,000,000,  N  =  784. 

10.  A  55o-kw.  generator  has  a  terminal  voltage  of  550  at  no-load  and  is  5 
per  cent  overcompounded.     Find:  (a)  the  size  of  wire  required  to  transmit  100 
amperes  a  distance  of  600  feet,  the  voltage  between  the  terminals  of  the  load 


78  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

apparatus  to  be  550,  (b)  the  watts  lost  in  the  line,  (c)  the  total  output  of  the 
generator. 

11.  The  rated  output  (line)  of  a  compound  generator  =150  amperes,  and 
its  no-load  voltage  is  220.    The  current  is  transmitted  over  a  line,  the  resistance 
of  which  is  0.166  ohm,  and  the  voltage  at  the  load  terminals  is  the  same  at  full 
load  as  at  no-load.     Find:  (a)  the  terminal  voltage  of  the  generator  at  full  load, 
(b)  percentage  overcompounding,  (c)  watts  lost  in  line,  (d)  watts  output  of 
gen  erator. 

12.  A  4-pole,  wave-wound  armature   has    105   commutator  bars.     Each 
armature  coil  consists  of  two  turns  of  No.  10  double  cotton-covered  wire.     Find 
the  flux  (per  pole)  required  when  the  induced  voltage  is  125,  and  the  speed 
=  looo  r.p.m. 

13.  A  4-pole,  lap- wound  armature  has  88  commutator  bars.    Each  arma- 
ture coil  consists  of  four  turns  of  No.  1 1  double  cotton-covered  wire.     Find  the 
voltage  when  the  flux  per  pole  =  1,000,000  lines  and  the  speed  of  the  armature 
=  1000  r.p.m. 


CHAPTER  V 
THE  CONTINUOUS-CURRENT  MOTOR 

i.  The  fundamental  equation  of  the  motor.  —  Rotation  of  the 
armature  of  a  motor  induces  in  the  armature  winding  a  counter- 
electromotive  force  which  is  dependent  on  the  same  quantities  as 
is  the  induced  electromotive  force  of  a  generator. 

Counter  e.m.f.  =    ,     \  volts,  (i) 

p  icr 

when  <£  =  the  total  flux  entering  or  leaving  the  armature  at  each 

pole, 
N  =  the  total  number  of  conductors  on  the  surface  of  the 

armature, 

n  =  the  speed  of  the  armature  (revolutions  per  second), 
p  =  the  number  of  poles, 
pr  —  the  number  of  parallel  paths  into  which  the  armature 

conductors  are  connected. 

The  counter-electromotive  force  is  the  difference  between  the 
applied  electromotive  force  and  the  resistance  drop  in  the  armature 
circuit.  Therefore, 

<f>Nn 


,  . 

E  "  RJa    Eb'  (2) 

when     E  =  the  applied  electromotive  force, 

Ra  =  the  resistance  of  the  armature  winding, 
Ia  =  the  current  flowing  hi  the  armature  circuit, 
Eb  =  the  drop  due  to  the  so-called  contact  resistance  be- 
tween the  brushes  and  the  commutator.* 

*  As  noted  in  Chapter  4,  Section  2,  the  voltage  drop  per  pair  of  carbon  brushes  may 
be  calculated  by  the  formula: 

Eb  =»  0.8  +  0.2  D,  (3) 

when  D  —  the  current  density  (amperes  per  square  centimeter)  in  the  contact  area 
between  the  brush  and  the  commutator. 

79 


8o  ESSENTIALS  .OF  ELECTRICAL  ENGINEERING 

Transposing  the  quantities  in  equation  (2), 


kEc 


since  />,  pr  and  N  are  constant  for  any  given  motor.  Equation  (4) 
is  the  usual  form  of  the  fundamental  equation  for  the  continuous- 
current  motor. 

2.  Torque.  —  The  force  tending  to  rotate  the  armature  of  a  motor 
is  termed  its  torque  and  is  usually  expressed  in  pounds  at  one  foot 
radius,  i.e.,  in  foot-pounds.  It  is  proportional  to  the  product  of 
the  flux  in  the  air  gap  and  the  current  flowing  in  the  armature 
circuit.  Let 

T  =  the  torque  in  foot-pounds  developed  in  the  armature  of  any 

motor, 
n  =  the  speed  (revolutions  per.  second)  at  which  the  armature 

rotates, 
EC  =  the  counter-electromotive  force  (volts)  induced  in  the  arma- 

ture conductors, 
Ia  =  the  current  (amperes)  flowing  in  the  armature  circuit. 

Then          MI.BJ.  (5) 

,  , 


and  T= 

'         8 

The  quantity 


is  constant  for  any  given  motor. 

The  torque  delivered  at  the  pulley  of  the  motor  is  less  than  that 
indicated  by  equation  (7),  because  part  of  the  torque  developed  in 
the  armature  is  required  to  overcome  the  windage,  friction  and  iron 
losses  (stray  power)  of  the  motor  itself. 

3.  Speed-torque  characteristic.  —  The  operation  of  a  motor  is 
represented  graphically  by  a  speed-torque  curve,  which  shows  the 


THE   CONTINUOUS-CURRENT  MOTOR  8l 

relation  between  the  speed  of  the  armature  and  the  torque  developed 
in  the  armature,  or  that  delivered  at  the  pulley. 

The  relations  between  the  speed,  the  torque,  and  the  armature 
current  of  a  continuous-current  motor  are  such  that  a  change  in  one 
of  these  quantities  produces  a  change  in  one  or  both  of  the  others. 
In  any  given  motor,  the  speed  is  automatically  adjusted  to  the 
value  which  allows  the  required  current  to  flow  in  the  armature 
conductors.  If  the  armature  current  is  larger  than  that  necessary 
to  produce  the  required  torque,  the  speed  of  the  armature  auto- 
matically increases  until  equilibrium  is  established;  -if  the  current 
is  too  small,  the  speed  of  the  armature  decreases,  allowing  a  larger 
current  to  flow  and  a  larger  torque  to  be  exerted. 

4.  Commutation.  —  The  phenomena  of  commutation  occur  in  a 
motor  just  as  in  a  generator,  since  the  direction  of  the  current  in 
each  armature  coil  is  periodically  reversed. 

5.  Armature  reaction.  —  The  direction  of  current  in  the  arma- 
ture conductors  of  a  motor  is  opposite,  for  a  given  field  polarity  and 
direction  of  armature  rotation,  to  that 

in  the  conductors  of  a  generator.  The 
field-  is,  therefore,  distorted  as  indi- 
cated in  Fig.  68,  and  the  brushes  must 
be  shifted  backward,  or  opposite  the 
direction  of  armature  rotation,  to  ob- 
tain good  commutation.  FlG>  68-  .Flux  Distribution  in  the 
s  „,.  iiT-,,1  Air  Gap  of  Motor. 

6.  The    shunt   motor.  —  With  con- 
stant applied  voltage,  the  field  current  and  the  armature  flux  of  a 
shunt  motor  are  constant,  neglecting  armature  reaction  which  is 
small  in  a  well-designed  motor,  and  the  torque  developed  in  the 
motor  is  proportional  to  the   current   flowing  in   the   armature 
circuit. 

T  oc  Ia.  (8) 

Characteristic.  —  As  the  load  on  a  shunt  motor  increases,  the 
counter-electromotive  force  must  decrease  so  that  the  required  cur- 
rent may  flow  in  the  armature  circuit.  Armature  reaction  reduces 
the  flux  of  a  shunt  motor  slightly,  but  if  the  magnetic  circuit  is 
properly  designed  this  reduction  is  small,  and  may  be  neglected.* 

*  Any  decrease  in  the  flux  of  a  shunt  motor  by  reason  of  armature  reaction,  tends 
to  increase  its  speed  and  improve  the  regulation  of  the  motor. 


82 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


The  speed  of  a  shunt  motor  must,  therefore,  decrease  as  the  load 
increases,  and  the  speed-torque  characteristic  is  a  slightly  drooping 
curve  as  indicated  in  Fig.  69. 

Well-designed  shunt  motors  operate  with  only  a  small  change  in 
speed  between  no-load  and  full  (rated)  load,  and  are  termed  con- 


20  40  60  &0  100         120 

PER  CENT  OF  FULL  LOAD( RATED) TORQUE. 

FIG.  69.    Speed-torque  Characteristic  of  Shunt  Motor. 


140 


stant-speed  motors.    They  are  largely  used  where  a  small  variation 
of  speed  is  of  little  consequence. 

Regulation.  —  The  regulation  of  a  shunt  motor  is  the  ratio  of  the 
difference  between  the  no-load  and  the  full-load  (rated)  speeds,  and 
the  full-load  speed. 


i  *• 
Per  cent  regulation  = 


—  n)  ioo 


n 


(9) 


when 


HQ  =  the  speed  at  no  load, 

n  =  the  speed  at  full  (rated)  load. 


The  resistance  of  the  armature  circuit  is  the  controlling  factor  in 
the  regulation  of  a  shunt  motor.  If  the  armature  resistance  is 
small,  the  variation  of  the  product,  RJa,  between  no  load  and  full 
load  is  small,  and  the  required  variation  of  the  counter-electromotive 
force  is  correspondingly  small.  But  the  counter-electromotive 
force  is  proportional  to  the  speed  of  the  armature.  Therefore,  a 
small  armature  resistance  produces,  or  tends  to  produce,  good 
regulation. 

Speed  control*  —  Speed  regulation  is  an  inherent  property  of  a 
motor;  speed  control  is  the  variation  in  speed  obtained  by  exter- 
nal means. 

The  speed  of  a  shunt  motor  is  varied,  within  certain  limits,  by: 
(a)  changing  the  resistance  of  the  armature  circuit,  (b)  changing  the 

*  For  an  extended  discussion  of  motor  speed  control,  see  "  Electric  Motors  "  by 
Crocker  and  Arendt. 


Field  Resistance 


THE  CONTINUOUS-CURRENT  MOTOR  83 

resistance  of  the  field  circuit,  (c)  changing  the  reluctance  of  the 
magnetic  circuit,  (d)  changing  the  electromotive  force  between  the 
armature  terminals. 

(a)  Changing  the  resistance  of  the  armature  circuit.  —  The  speed  of 
a  motor  is  changed  by  varying  a  resistance  connected  in  series  with 
the  armature  as  indicated  in  Fig.  70.     A 

resistance,  the  current-carrying  capacity  of 
which  is  equal  to  or  greater  than  the  maxi- 
mum current  that  will  flow  in  the  circuit, 
must  be  used.  This  method  is  effective, 
cheap  in  first  cost,  and  easily  applied,  but 

the  power  wasted  in  heating  the  rheostat  FlG'  7a    sPeed  Contro1  by 

Armature  Resistance, 
makes  it  very  inefficient.     As  pointed  out 

above,  a  small  change  in  load  causes  a  large  change  in  speed,  and 
the  motor  acts  as  if  the  resistance  of  the  armature  winding  were 
greatly  increased. 

(b)  Changing  the  resistance  oj  the  field  circuit.  —  The  resistance  of 
the  field  circuit  is  changed  by  manipulating  the  field  rheostat.     In- 
creasing the  resistance  of  the  field  circuit  decreases  the  field  cur- 
rent and  increases  the  speed;    decreasing  the  resistance  of  the 
field  circuit  increases  the  field  current  and  decreases  the  speed.* 

This  method  is  cheap,  both  as  to  first  cost  and  as  to  operation, 
but  its  range  of  operation  is  limited.  Very  low  speeds  are  unattain- 
able because  of  the  fact  that  above  a  certain  point  a  large  increase 
in  field  excitation  produces  only  a  small  increase  in  flux.  High 
speeds  are  also  unattainable  because,  with  a  weak  field,  armature 
reaction  becomes  so  pronounced  as  to  cause  excessive  sparking 
which  burns,  and  would  ultimately  destroy,  the  commutator. 

Armature  reaction  and  commutation  difficulties  are  materially 
reduced  by  means  of  interpoles  which  provide  a  local  commutat- 
ing  flux,  and  make  possible  the  operation  of  shunt  motors  having 
very  weak  fields.  Interpole  motors  having  a  maximum  speed  six 
times  the  minimum  speed,  and  controlled  by  manipulation  of  the 
field  rheostat,  are  in  successful  operation. 

(c)  Changing  the  reluctance  of  the  magnetic  circuit.  —  With  con- 
stant field  excitation,  the  flux  in  the  magnetic  circuit  of  a  dynamo  is 
inversely  proportional  to  the  reluctance  of  the  circuit.f    A  number 

*  This  statement  is  evident  from  equation  (4). 
t  See  Chapter  2,  Section  10. 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


Field  Resistance 


of  variable-speed  motors  have  been  designed  in  which  the  reluctance 
of  the  magnetic  circuit  is  varied  by  changing  the  length  of  the  air 
gap  between  the  pole  and  the  armature.  The  operation  of  such 
motors  is  satisfactory,  but  they  are  expensive  in  construction  and 
more  or  less  complicated  in  operation. 

(d)  Changing  the  electromotive  force  applied  to  the  armature  terminals. 
—  This  method  of  speed  control  may  be  sub-divided  into  two  parts: 
(i)  when  the  change  in  voltage  is  made  in  steps,  (2)  when  the  range 
of  voltage  is  continuous. 

(i)  Voltage  changed  in  steps.  —  This 
method  of  speed  control  necessitates  the 
use  of  several  supply  mains  having  dif- 
ferent voltages,  and  a  generating  system 
capable  of  supplying  current  at  these 
FIG.  71.  Multi-voltage  Speed  different  voltages.  The  field  winding 

is  permanently  connected  across  one  pair 

of  the  supply  mains  and  the  flux  is,  therefore,  approximately  con- 
stant. Fig.  71. 

The  operating  characteristics  of  this  system  are  good  and  the 
efficiency  high,  the  objections  being  its  high  first  cost  and  the  fact 
that  the  speed  changes  by  fixed  steps.  The  latter  fault  is  remedied 
by  combining  this  method  with  the  field  rheostat  method. 

•  (2)  Voltage  range  continuous.  —  This  method,  devised  by  Ward 
Leonard  and  generally  known  by  his  name,  requires  the  use  of 
three  machines  —  a  vari- 
able-speed motor,  a  con- 
stant-speed motor*  and 
a  generator,  as  shown  in 
Fig.  72.  The  constant- 
speed  motor  drives  the 
generator,  and  the  gen- 
erator, in  turn,  supplies 
current  to  the  armature  of 
the  variable-speed  motor, 
the  field  of  which  is  connected  to  constant  potential  mains.  By 
varying  the  field  excitation  of  the  generator,  any  desired  voltage  is 
delivered  at  the  terminals  of  the  motor  armature,  and  a  continuous 
variation  of  speed  obtained  over  the  widest  possible  range.  Very 
*  It  is  not  necessary  that  this  be  an  electric  motor. 


Generator 


Constant  speed  Hohr 


Variable  Speed  Motor 

FIG.  72.     Ward  Leonard  Speed  Control  System. 


THE   CONTINUOUS-CURRENT  MOTOR 


simple  operating  mechanism  is  required.  The  efficiency  of  this 
system  is  the  combined  efficiency  of  three  machines,  the  individual 
efficiencies  of  which  are  usually  low.  Its  high  first  cost  and  low 
operating  efficiency  prohibit  its  use  except  where  these  factors  are 
secondary  considerations. 

7.  The  series  motor.  —  The  series  motor,  unlike  the  series 
generator,  has  a  very  large  commercial  application.  It  is  used, 
practically  to  the  exclusion  of  all  other  types  of  motor,*  for  street 


180 
i  fifi 

i 

250 
225 

H- 

z 
uJ 

200  or 

K 

3 

0 
U 

I25< 
100^ 

u- 

75  1 

UJ 

UJ 
CL 

25 

» 

\ 

\ 

\ 

t 

f 

140 
100 

ao 

60 
n 

x4 

^ 

J 

rl 

^peed 

f 

\ 

x^ 

\ 

X1 

\ 

^, 

/ 

\ 

,/ 

'  ( 

sU 

r/ 

-o 

-jf 

\ 

^M 

| 

\ 

X1 

•*, 

/ 

<^ 

•s^ 

. 

^x 

's 

^ 

X 

"S^ 

~j 

^ 

jj 

/ 

^ 

^N 

^ 

v~« 

•^^ 

^ 

/ 

*, 

a 

/ 

t 

/ 

/ 

40  80  120  160          200        340 

PER  CENT  OF  FULL  LOAD  (RATED)  TO  ROUE 

FIG.  73.    Series  Motor  Curves. 


280 


railway  work,  for  hoisting  and  for  all  purposes  which  require  large 
starting  torques,  and  do  not  require  close  speed  regulation. 

Torque.  —  Since  the  field  excitation  of  a  series  motor  is  propor- 
tional to  the  armature  current,  the  torque  developed  in  its  armature 
is,  theoretically,  proportional  to  the  square  of  the  armature  current. 

roc/a2.  (10) 

This  equation  holds  good  for  low  values  of  armature  current,  but  the 
upper  part  of  the  current-torque  curve  is  found  to  be  an  approxi- 
mately straight  line.     Fig.  73.     For  low  excitations,  the  flux  ispro- 
*  Alternating-current  motors  are  used  to  a  limited  extent  for  these  purposes. 


86 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


portional  to  the  field  current  but  for  larger  loads  the  iron  of 
the  magnetic  circuit  approaches  saturation,  and  the  flux  in- 
creases more  slowly  than  does  the  current  in  the  field  winding. 
The  torque  developed  is,  therefore,  less  than  that  indicated  by 
equation  (10). 

Speed.  —  The  speed  of  a  series  motor  varies  over  wide  limits  as 
the  load  changes,  the  speed  increasing  as  the  load  is  reduced  as  indi- 
cated in  Fig.  73.  The  full  line  shows  the  operating  range  of  the 
motor,  the  dotted  portion  near  the  axis  of  ordinates  indicates  speeds 
above  the  safe  operating  limits,  and  excessive  heating  takes  place 
if  the  motor  is  operated  over  the  dotted  portion  at  the  right. 

Because  of  the  excessive  speed  attained  by  the  armature  of  a 
series  motor  when  the  load  is  small,  a  series  motor  should  never 
be  used  where  the  load  may,  accidentally  or  otherwise,  be  reduced 
below  the  safe  minimum  value,  or  the  motor  may  be  wrecked. 


PER  CENT  OF  FULL  LOAD 
(RATED)  SPEED 
o  o  —  f^ 

0  0  0  < 

'  —  , 

»^ 

—  ^. 

x_ 

•^ta 

1 

T/ 

"- 

"^ 

\  L 

§ 

\ 

& 

& 

/ 

N 

<J 

/ 

*J"1 

-« 

i  i 

A 

differential 

I    |        ]S 

\/. 

^ 

i  4-*f"xi 

1 

33* 

\ 

i     i  i  i 

40  60  60          100          120 

PER  CENT  OF  FULL  LOAD  (RATED)  TORQUE 

FIG.  74.    Speed-torque~Curve~of  Compound  Motors. 


140 


8.  Compound  motors.  —  Compound  motors  are  divided  into 
two  classes:  (a)  cumulative,  (b)  differential. 

(a)  Cumulative  compound  motor.  —  The  shunt  and  the  series-field 
windings  of  a  cumulative  compound  motor  are  so  connected  as  to 
produce  fluxes  in  the  same  direction.  Therefore,  the  excitation  of 
the  motor  increases  with  the  load,  and  the  speed  of  the  armature 
may  be  materially  reduced.  The  shape  of  the  speed-torque 
curve  of  a  cumulative  compound  motor  depends  on  the  relative 
effects  of  the  two  field  windings,  but  that  part  of  the  curve  repre- 
senting light  loads  is  usually  similar  to  that  of  a  shunt  motor  with 
large  armature  resistance.  As  the  load  increases,  the  characteristic 
gradually  reverses  and  finally  approximates  that  of  a  series  motor. 
Fig.  74  shows  the  speed-torque  characteristic  of  a  cumulative  com- 
pound motor. 


THE   CONTINUOUS-CURRENT  MOTOR  87 

A  cumulative  compound  motor  having  a  heavy  flywheel  is  well 
adapted  for  the  operation  of  shears,  punches  and  other  apparatus 
where  the  load  is  applied  suddenly.  During  the  periods  of  no  load, 
energy  is  stored  in  the  flywheel  and  other  rotating  parts.  When 
the  load  is  applied,  this  stored  energy  is  returned  to  the  system  and 
prevents  the  speed  of  the  motor  from  dropping  to  as  low  a  value  as 
would  otherwise  be  the  case,  and  reduces  the  range  over  which  the 
armature  current  varies. 

(b)  Differential  compound  motor.  —  The  field  windings  of  a  dif- 
ferential compound  motor  oppose  each  other  so  that  the  total  flux 
in  the  magnetic  circuit  decreases  as  the  armature  current  increases. 
The  speed  of  a  differential  compound  motor  is  constant  if  the  ratio 
of  the  counter-electromotive  force  and  the  flux  is  constant. 

By  properly  designing  the  magnetic  circuit,  the  speed  of  a  differ- 
ential compound  motor  is  maintained  practically  constant  between 
no  load  and  full  (rated)  load,  as  indicated  in  Fig.  74.  Over  that 
portion  of  the  curve  where  the  speed  is  constant,  the  flux  decreases 
slightly,  and  the  torque  is  approximately  proportional  to  the  arma- 
ture current.  When  the  speed  begins  to  rise  the  flux  decreases 
rapidly  and  soon  reaches  the  point  where  it  decreases  at  approxi- 
mately the  same  rate  as  the  armature  current  increases.  Any 
further  increase  in  armature  current  cannot  increase  the  torque, 
which  is  proportional  to  the  product  of  flux  and  armature  current, 
since  the  combined  effect  of  the  differential  winding  and  armature 
reaction  reduces  the  flux  faster  than  the  armature  current  increases. 
Excessive  weakening  of  the  field  causes  poor  commutation  and  is 
indicated  by  sparking. 

From  the  above,  it  is  evident  that  a  differential  compound  motor 
cannot  be  greatly  overloaded.  Also,  because  of  the  large  current 
which  would  flow  in  the  series  winding  during  the  starting  period 
and  largely  neutralize  the  effect  of  the  shunt-field  winding  so  that 
the  starting  torque  would  be  small,  the  series  winding  is  usually 
short-circuited  during  the  starting  period,  i.e.,  the  motor  is  started 
as  a  shunt  motor.  The  commercial  application  of  the  differential 
compound  motor  is  very  limited. 

9.  Motor-starting  rheostats.  —  Because  of  the  very  low  resistance 
of  the  armature  winding  of  a  motor,  an  excessive  current  will  flow 
if  the  circuit  is  connected  directly  to  the  supply  mains  to  start  it. 
It  is,  therefore,  necessary  to  connect  an  external  resistance  in  series 


88 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


with  the  armature  winding  during  the  period  of  acceleration.  As 
the  speed  of  the  armature  increases,  the  counter-electromotive  force 
opposing  the  flow  of  the  current  increases,  and  the  resistance  is 
gradually  reduced  until  the  armature  terminals  are  connected 
directly  to  the  supply  mains. 

Rheostats  for  shunt  and  compound  motors.  —  The  starting  rheo- 
stats used  on  shunt  and  compound  motors  are  of  various  designs, 


voltage 


Low  voltage 
release 


FIG.  75.     Motor  Starting  Rheostat. 
General  Electric  Co. 


FIG.  76.     Motor  Starting  Rheostat. 
General  Electric  Co. 


typical  examples  being  shown  in  Figs.  75  and  76.     The  internal  and 
external  electrical  connections  are  indicated  in  Figs.  78  and  79. 

The  handle  H  (Fig.  78)  when  moved  so  as  to  make  contact  with 
stud  i,  closes  the  armature  circuit  through  the  resistance  R.     This 

movement  of  the  handle  H  also 
closes  the  field  circuit.  As  the 
speed  of  the  armature  increases, 
the  handle  H  is  moved  to  studs 
2,  3,  etc.,  until  the  entire  resistance 
is  cut  out. 

The  field  circuit  includes  the 
exciting  coil  of  an  electromagnet 
which  holds  the  handle  H,  against 
the  pull  of  a  spring,  in  contact 
with  the  last  stud.  Should  the 
supply  circuit  be  opened  or  the  field 
excitation,  for  any  reason,  be- 
come greatly  weakened,  the  magnet 
can  no  longer  hold  the  handle,  the  spring  returns  it  to  the  posi- 
tion indicated  in  the  figure  and  opens  the  armature  circuit.  This 
magnet  is  known  as  the  "  low- voltage "  release,  sometimes  errone- 
ously called  the  no-load  release,  and  prevents  injury  to  the  appa- 
ratus should  the  supply  be  temporarily  interrupted. 


FIG.  77.  Motor  Starting  Rheostat 
(Back  and  Side  Removed).  West- 
inghouse  Elec.  &  Mfg.  Co. 


THE   CONTINUOUS-CURRENT  MOTOR 


89 


The  winding  of  the  other  electromagnet  shown  in  Fig.  78  is  con- 
nected in  series  with  the  armature  and  so  arranged  that  at  some 
predetermined  current  value,  switch  5  is  closed,  and  the  winding 
of  the  low- voltage  release  short-circuited.  The  low- voltage  release 
is  thus  deprived  of  its  magnetizing  current  and  the  handle  returns 
to  the  open-circuit  position,  the  supply  circuit  is  opened,  and  the 
motor  stops.  This  device  is  the  "overload"  release  which  pro- 


Starting  Resistance 


Armahirr 

FIG.  78.     Wiring  Diagram  for  Shunt      FIG.  79.     Wiring  Diagram  for  Shunt 
Motor  and  Starting  Rheostat.  Motor  and  Starting  Rheostat. 

tects  the  armature  from  excessive  currents.  Circuit  breakers  con- 
nected in  the  supply  lines  accomplish  the  same  purpose  and  are  to 
be  preferred,  for  when  the  circuit  is  opened 
by  means  of  the  over-load  release,  the  studs 
of  the  starting  rheostat  are  burned,  becom- 
ing rough  and  irregular. 

Fig.  79  is  similar  to  Fig.  78  except  the 
overload  release  is  omitted,  and  the  low- 
voltage  release  is  connected  directly  across 
the  supply  circuit. 

Rheostats  for  series  motors. — The  starting 
rheostats  used  with  series  motors  are  usually 
of  the  drum  type.  Fig.  80.  Low- voltage 
releases  are  not  essential  on  series  motors 
as  an  attendant  is  usually  near  to  open  the  FlG-  8o-  Drum  Controller, 
circuit  and  return  the  handle  to  the  "off" 

position  in  case  the  current  supply  fails.  Circuit  breakers,  which 
should  be  connected  in  the  supply  leads  of  each  motor,  may  be 
depended  on  to  protect  the  armature  from  excessive  currents. 


9o 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


The  series-parallel  system  of  motor  control  is  used  on  street  cars, 
which  have  two  or  more  motors,  start  and  stop  frequently,  and 
are  often  required  to  run  at  low  speeds.  Its  advantage  is  the  re- 
duction of  the  losses  in  the  starting  rheostat  and  a  correspondingly 
increased  operating  efficiency. 

For  starting,  the  two  motors  are  connected  in  series  with  each 
other  and  with  the  starting  resistance,  which  makes  it  possible  to 
use  a  smaller  resistance  than  would  otherwise  be  required.  Fig.  8ia. 
As  the  armature  speed  increases,  the  resistance  is  reduced  as  in 

any  other  starting  system. 
After  all  the  starting  re- 
sistance has  been  cut  out 
of  the  circuit  and  the 
motors  have  attained  their 
maximum  speed  with  the 
series  connection  (one-half 
the  line  voltage  applied  to 
each  motor),  the  motors 

are    connected  in  parallel 
FIG.  81.    Series-parallel  Motor  Control.  .^  ^       ^  ,     . 

series  with  the  starting  resistance.  Fig.  8ib.  As  the  motors  speed 
up  further,  the  external  resistance  is  reduced  until  each  motor 
is  connected  directly  across  the  supply  lines.  With  this  arrange- 
ment it  is  evident  that  the  motors  may  be  operated  at  two  different 
speeds  without  any  loss  due  to  an  external  resistance. 

All  the  operations  required  in  the  series-parallel  system  of  control 
are  obtained  by  moving  a  handle  which  rotates  a  spindle  having 
metal  fingers  or  sectors.  The  same  general  type  and  arrangement 
of  rheostat  is  used  for  crane  and  other  series  motors,  although  only 
one  motor  may  be  used. 

Automatic  motor-starting  rheostats.  —  Motor-starting  rheostats  in 
which  the  resistance  of  the  armature  circuit  is  automatically  re- 
duced as  the  speed  of  the  armature  increases,  have  been  developed. 
With  such  apparatus  it  is  only  necessary  to  close  or  open  a  line 
switch  to  start  or  stop  the  motor. 


(b) 


(a) 


THE   CONTINUOUS-CURRENT  MOTOR  91 

CHAPTER  V  — PROBLEMS 

1.  A  current-carrying  wire  lies  in  a  magnetic  field,  the  intensity  of  which  is 
50,000.     Find  the  force  acting  on  the  wire  when  250  amperes  flow. 

2.  A  4-pole,  lap-wound,  22o-volt  shunt  motor  has  an  armature  input  of  200 
amperes  when  running  at  750  r.p.m. 

Diameter  of  armature 15  inches 

Length  of  armature 9  inches 

Number  of  armature  conductors 476 

Pole  enclosure 70% 

Armature  resistance 0.05 

Flux  density  on  pole  face  (average) 48,000 

Find:  (a)  the  torque  developed  in  the  armature,  (b)  the  speed  of  the  motor  at  no 
load  (no-load  armature  current  =  6.5  amperes),  (c)  the  speed  regulation,  assum- 
ing 200  amperes  to  be  rated  full  load. 

3.  Find  the  external  resistance  which  must  be  connected  in  series  with  the 
armature  of  the  motor  in  Problem  2  to  reduce  the  speed  to  400  r.p.m.,  the  arma- 
ture current  and  the  field  excitation  being  as  given  above. 

4.  Find  the  speed  of  the  armature  of  the  motor  in  Problem  2  when  connected 
in  series  with  the  resistance  required  in  Problem  3,  with  100  amperes  flowing  in 
the  armature  circuit. 

5.  Calculate  the  per  cent  of  total  power  supplied  to  the  motor  dissipated  in 
heating  the  rheostats  in:  (a)  Problem  3,  (b)  Problem  4. 

6.  Find  the  resistance  of  the  starting  rheostat  required  for  the  motor  in 
Problem  2,  the  maximum  current  during  the  period  of  acceleration  not  to  ex- 
ceed 150  per  cent  of  the  rated  full-load  current. 

7.  The  normal  field  current  of  a  shunt  motor  (temperature  of  shunt  field 
=  60°  C.)  is  10  amperes.    Find  the  field  current  when  first  started,  assuming 
the  temperature  of  the  air  to  be  20°  C. 

8.  Two  similar  15  horse-power  22o-volt  shunt  motors  are  connected  in  series 
between  44o-volt  mains,  the  armatures  of  the  motors  being  rigidly  connected 
together.     Resistance  of  each  armature  =  0.08  ohm.    Armature  current  =  60 
amperes.    Find  the  speed  at  which  the  armatures  rotate  when:  (a)  the  fields 
are  normally  (and  equally)  excited,  (6)  the  field  excitations  are  such  that  the 
flux  density  in  the  air  gap  of  one  motor  is  25  per  cent  greater  than  that  in  the 
air  gap  of  the  other. 

9.  A  shunt  dynamo  when  operated  as  a  generator  at  1000  r.p.m.  delivers  100 
amperes  (armature)  current  at  a  terminal  voltage  of  220.    The  resistance  of  the 
armature  circuit  is  0.05  ohm.    Calculate  the  speed  at  which  the  armature 
rotates  if  operated  as  a  motor,  the  armature  current  and  the  field  excitation 
being  the  same  as  in  the  generator. 

10.  A  4-pole,  wave- wound  shunt  motor  is  operated  on  a  2So-volt  circuit. 
Armature  slots  =  47,  conductors  per  slot  =12,  commutator  bars  =  141,  flux 
=  1,660,000  maxwells  per  pole.    Neglecting  armature  resistance,  find  the  speed 
of  the  motor. 


92  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

11.  The  armature  of  a  6-pole  continuous-current  motor  is  10  inches  long 
and  24  inches  in  diameter.     The  face  of  each  pole  shoe  is  10  inches  X  n  inches. 
The  armature  is  lap  wound  and  consists  of  600  conductors.     The  average  flux 
density  on  the  face  of  the  poles  is  48,000  lines  per  square  inch.     Find  the  force 
acting  on  the  armature  when  the  total  input  to  the  armature  is  300  amperes. 

12.  A  230- volt  shunt  motor  has  a  no-load  input  of  30  amperes,  and  a  full- 
load  input  of  750  amperes  when  operated  at  400  r.p.m.     Armature  resistance 
=  0.0075;  field  resistance  =  46.     Find  the  speed  regulation  of  the  motor. 

13.  Same  as  Problem  12  except  the  armature  resistance  =  0.015. 

14.  The  armature  of  the  motor  in  Problem  12  is  6-pole,  lap  wound,  and  has 
540  conductors.     Find  the  flux  per  pole. 


CHAPTER  VI 


LOSSES,   EFFICIENCIES   AND   RATINGS   OF   CONTINU- 
OUS-CURRENT DYNAMOS 

i.  Losses.  —  The  losses  in  a  continuous-current  dynamo  are: 
(a)  resistance  losses,  (b)  stray  power. 

(a)  Resistance  losses*  —  The  resistance  losses  of  a  dynamo  are 
those  due  to  the  resistance  of:  (i)  the  armature  winding,  (2)  the 
brush  contacts,  (3)  the  field  circuit. 

(i)  Armature  resistance  losses.  —  Armature  resistance  losses  are 
not  measured  directly,  but  are  calculated  from  the  resistance  of  the 
armature  winding  and  the  given  or  required  value  of  armature 
current. 


Pa  = 


/  \ 

(i) 


when 


Ra  =  the  resistance  of  the  armature  winding, 
/«  =  the  armature  current  for  which  the  loss  is  to  be  calcu- 
lated. 


Connections  for  the  determination  of  the  resistance  of  the  arma- 
ture winding  are  shown  in  Fig.  82. 
After  opening  the  field  circuit  so  that 
the  armature  will  not  rotate,  the 
armature  is  connected  to  supply  mains 
in  series  with  a  water  rheostat  or 
other  suitable  resistance,  by  means  of 
which  the  current  in  the  armature  FlG-  82-  Connections  for  Deter- 
circuit  may  be  controlled.  The  cur-  ***  c 


rent  in  the  armature  circuit  and  the 

electromotive  force  between  brush  contacts  are  measured  by  means 

of  the  ammeter  and  the  voltmeter. 

Then  ^  =  |,  (2) 

when  Ra  =  the  resistance  of  the  armature  winding, 
E  =  the  indication  of  the  voltmeter, 
/  =  the  current  flowing  in  the  armature  circuit. 
*  Resistance  losses  should  be  calculated  for  a  working  temperature  of  75  °  C. 

93 


94  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

(2)  Brush-contact  losses.  — As  indicated  in  Chapter  4,  Section  2, 
the  so-called  brush-contact  resistance  is  not  constant,  but  is  a  func- 
tion of  the  current  density  in  the  contact  area,  and  the  drop  per 
pair  of  carbon  brushes  is  calculated  from  the  formula, 

Eb  =  0.8  +  0.2  D,  (3) 

when  D  =  the  current  density  (amperes  per  square  centimeter)  in 
the  contact  area  between  the  brush  and  the  commu-- 
tator. 

Therefore  the  loss  due  to  brush-contact  resistance  is 

Pb  =  (0.8  +  0.2  D)Ia  watts.  (4) 

(3)  Field  resistance  losses.  —  The  resistance  of  a  series-field  wind- 
ing is  small  and  should  be  determined  in  the  same  way  as  that  of 
an  armature  winding.     The  copper  loss  may  then  be  calculated. 

P8  =  RJ*  watts,  (5) 

when     Rs  =  the  resistance  of  the  series-field  circuit, 

I8  =  the  current  flowing  in  the  series-field  circuit. 

The  resistance  of  a  shunt-field  circuit  is  comparatively  large  and 
the  winding  may  be  connected  directly  to  the  supply  mains,  pro- 
vided the  voltage  of  the  line  is  not  materially  greater  than  the  rated 
voltage  of  the  dynamo.  The  losses  in  the  shunt-field  circuit,  in- 
cluding the  rheostat,  may  be  determined  by  direct  measurement,  or 
by  calculation. 

Pf  =  EIf  watts  (6) 

=  Rfl?  watts      ,      i  ,    ,  (7) 

~f  watts  (8) 

&t 

when    E  =  the  voltage  between  the  terminals  of  the  field  circuit, 
If  =  the  current  flowing  in  the  field  circuit, 
Rf  =  the  resistance  of  the  field  circuit. 

The  instruments  required  for  the  determination  of  either  the  field 
loss  or  the  resistance  of  the  circuit,  are  an  ammeter  and  a  voltmeter. 

(b)  Stray  power.  —  The  stray  power  of  a  dynamo  includes  all  the 
losses  not  specified  in  (a),  and  consists  of:  (i)  iron  losses,  (2)  fric- 
tional  losses. 


LOSSES,  EFFICIENCIES  AND  RATINGS  95 

(i)  Iron  losses.  —  The  iron  losses  of  a  dynamo  are  due  to  hys- 
teresis and  eddy  currents,  and  are  dependent  on  the  speed  at 
which  the  armature  rotates,  and  on  the  flux  density  in  the  arma- 
ture core. 

Hysteresis.  —  Since  the  armature  rotates  in  a  magnetic  field,  the 
magnetism  of  the  iron  core  is  periodically  reversed,  and  this  re- 
arrangement of  the  molecular  structure  of  the  iron  requires  an 
expenditure  of  energy  which  heats  the  iron.  The  energy  required 
to  reverse  the  magnetic  polarity  of  any  volume  of  iron  has  been 
found,  experimentally,  to  be  approximately  proportional  to  the  1.6 
power  *  of  the  maximum  flux  density  in  the  iron.  The  power  lost 
by  hysteresis  is,  then,  proportional  to  the  1.6  power  of  the  flux 
density  in  the  armature  core,  and  to  the  number  of  magnetic  re- 
versals per  unit  of  time,  i.e.,  to  the  speed  of  the  armature. 

Ph  =  khnB™  watts,f  (9) 

when    kh  =  a  constant  dependent  on  the  volume  and  the  magnetic 
quality  of  the  iron  in  the  armature  core,  and  on  the 
number  of  poles  on  the  dynamo, 
n  —  the  speed  of  the  armature, 
B  =  the  flux  density  in  the  iron. 

Eddy  currents.  —  When  the  armature  rotates  in  the  magnetic  field 
set  up  by  the  field  windings,  electromotive  forces  are  induced  which 

*  See  Appendix  B,  Section  7. 

t  Professor  Sheldon  gives  the  following  formulae  for  the  calculation  of  hysteresis 
and  eddy-current  losses  in  iron: 

Ph  =  8.3  rjfVB1 8  lo-8  watts,  (9a) 

Pe  =  4.07  V  (JIB)*  io-17  watts,  (i2a) 

when        17  =  the  magnetic  (hysteretic)  constant, 

/  =  the  number  of  magnetic  reversals  per  second 


(_  number  of  poles  X  r.p.sA 
'   *•-.  / 


/  =  thickness  of  laminations  (in  mils), 
V  =  the  volume  of  iron  (in  cubic  inches), 
B  =  the  flux  density  in  the  iron  (maxwells  per  square  inch). 

Because  the  laminations  are  imperfectly  insulated  from  each  other  and  the  flux 
distribution  in  the  teeth  and  core  is  not  uniform,  the  calculation  of  hysteresis  and 
eddy-current  losses  in  armature  cores  is  only  a  rough  approximation.  Measured  values 
may  be  found  to  be  several  times  the  calculated  values. 


96 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


cause  currents  to  circulate  in  the  body  of  the  core.  Fig.  83.  These 
currents  cause  the  core  to  heat,  and  the  energy  lost  by  reason  of  their 
circulation  is,  by  Joule's  Law,  proportional  to  the  product  of  the 
square  of  the  current  and  the  resistance  of  the  paths  through  which 
they  flow. 

Referring  to  Fig.  83,  let 

n  =  the  speed  of  the  armature  (revo- 

lutions per  second), 
B  =  the  flux  density  of  the  magnetic 

field, 

/  =  the  thickness  of  the  laminations 
of  which  the  armature  is 
built, 

r  =  the  radial  distance  of  an  ele- 
ment parallel  to  the  axis  of 

^  armature   core    from   the 

axis  of  the  armature. 

k  and  ke  =  constants  proportional  to  the  volume  and  the  electri- 
cal quality  of  the  iron  in  the  armature  core,  and  to  the 
number  of  poles  on  the  dynamo. 

The  electromotive  force  generated  in  an  element  of  the  lamination 
due  to  the  rotation  of  the  core  in  the  magnetic  field  is 

e  =  2  irrlnB  io~8  volts.  (10) 

If  the  resistance  of  the  eddy  current  path  is  constant,*  the  losses 
due  to  the  circulation  of  the  eddy  current  in  the  iron  are  propor- 
tional to  the  square  of  the  induced  electromotive  force. 

Pe  =  ke*  (n) 

(12) 


FIG.  83.     Eddy  Currents  in  Solid 
Armature  Core. 


*  An  examination  of  the  eddy-current  path  shows  that  the  length  of  the  path  does  not 
vary  appreciably  for  the  different  thicknesses  of  laminations  used  in  commercial  dy- 
namos. 

t  Professor  Sheldon  gives  the  following  formulae  for  the  calculation  of  hysteresis 
and  eddy-current  losses  in  iron: 

ph  =  8.3  77/751'6  io~8  watts,  (ga) 

pe  =  4.07  V  (flB)z  io~17  watts,  (i2a) 

when        77  =  the  magnetic  (hysteretic)  constant, 

/  =  the  number  of  magnetic  reversals  per  second 
/  _  number  of  poles  X  r.p.s.\ 


LOSSES,  EFFICIENCIES  AND  RATINGS  97 

i.e.,  eddy-current  losses  are  proportional  to  the  square  of  the  thick- 
ness of  the  laminations,  to  the  square  of  the  armature  speed,  and  to 
the  square  of  the  density  of  the  magnetic  field. 

In  dynamos  having  toothed  armatures,  the  flux  density  on  the 
pole  face  is  not  uniform  but  is  greater  opposite  a  tooth  than  oppo- 
site a  slot.  As  the  armature  rotates,  each  element  of  the  pole  face 
parallel  to  the  axis  of  the  armature  shaft  has  an  electromotive  force 
induced  in  it  and  eddy-currents  circulate  in  the  iron.  Pole-face 
losses  are  reduced  by  laminating  the  poles  and  by  making  the 

width  of  slot 

ratio  —  -  small, 

length  of  air  gap 

(2)  Frictional  losses.  —  The  frictional  losses  of  a  dynamo  are 
those  between  the  shaft  and  the  bearings,  between  the  brushes  and 
the  commutator,  and  between  the  air  and  the  rotating  parts. 

Bearing  friction  and  windage  are  seldom  or  never  separated,  but 
in  a  well-designed  machine,  the  windage  is  small  and  may  be  neg- 
lected if  the  peripheral  speed  of  the  armature  does  not  exceed 
6000  feet  per  minute.  This  speed  is  seldom  exceeded  in  continu- 
ous current  machines  unless  they  are  to  be  direct  connected  to 
steam  turbines.  For  bearings  having  ring  lubrication  and  using 
light  machine  oil,  a  film  of  oil  always  separates  the  bushing  and  the 
shaft,  and  bearing  friction  may  be  assumed  to  be  fluid  friction. 
With  this  assumption,  the  loss  due  to  bearing  friction  is  calculated 
by  means  of  the  following  empirical  formula,  which  gives  close 
approximations  for  the  usual  range  of  armature  speeds. 

Pb  =  0.81  DL  ( -  -  }  watts,  (13) 

\IOO/ 

when    D  =  the  diameter  of  the  bearing  (in  inches), 
L  =  the  length  of  the  bearing  (in  inches), 
V  =  the  velocity  of  the  rubbing  surface  (in  feet  per  minute). 

The  above  formula  is  independent  of  pressure  and,  therefore,  of 
the  load  on  the  dynamo.  Bearing  friction,  in  belted  machines,  may 


/  =  thickness  of  laminations  (in  mils), 
V  =  the  volume  of  iron  (in  cubic  inches), 
B  =  the  flux  density  in  the  iron  (maxwells  per  square  inch). 

Because  the  laminations  are  imperfectly  insulated  from  each  other,  and  the  flux 
distribution  in  the  teeth  and  core  is  not  uniform,  the  calculation  of  hysteresis  and 
eddy-current  losses  in  armature  cores  is  only  a  rough  approximation.  Measured  values 
may  be  found  to  be  several  times  the  calculated  values. 


98  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

vary  over  wide  limits,  depending  on  the  tension  of  the  belt,  but 
the  losses  due  to  belt  tension  are  hard  to  estimate  and  are  usually 
neglected  in  making  efficiency  calculations. 

Brush  friction  depends  on  the  force  with  which  the  brushes  press 
against  the  commutator,  and  usually  amounts  to  only  a  small  part 
of  the  losses  in  a  dynamo.  A  pressure  of  one  and  one-half  to  two 
pounds  is  an  average  brush  pressure.  For  constant  pressure  be- 
tween the  brushes  and  the  commutator,  the  losses  due  to  brush 
friction  are  proportional  to  the  peripheral  velocity  of  the  commu- 
tator and  may  be  calculated  by  the  following  empirical  formula: 

PC  =  ^  watts,  (14) 

when    A  =  the  area  (in  square  inches)  of  the  brushes  in  contact 

with  the  commutator, 

V  =  the  peripheral  velocity  (in  feet  per   minute)  of    the 
commutator. 

From  the  above  considerations  it  is  evident  that  no  simple  equa- 
tion is  available  for  the  calculation  of  stray  power.  These  losses 
are,  therefore,  determined  experimentally  by  running  the  dynamo 
under  the  required  conditions  of  speed  and  field  excitation.  While 
the  separation  of  stray  power  into  its  components  is  not  a  difficult 
problem,  it  is  one  which  is  primarily  of  interest  to  the  designer  and 
the  manufacturer  only.  The  operating  engineer  is  interested  in 
obtaining  high  efficiency  and  good  operating  characteristics. 

2.  Experimental  determination  of  stray  power.  —  The  stray 
power  of  a  dynamo  is  easily  determined  by:  (a)  running  it  as  a 
motor,  (b)  driving  it  as  a  generator  by  means  of  an  auxiliary  motor. 

(a)  Running  as  a  motor.  —  When  a  dynamo  is  operated  as  a  motor 
without  load,  the  input  is  just  sufficient  to  supply  the  losses  —  field 
copper  loss,  armature  copper  loss,  brush-contact  resistance  loss  and 
stray  power.  The  resistance  losses  are  readily  determined  as  indi- 
cated above,  and  the  stray  power  is  the  difference  between  the  total 
input  (the  product  of  applied  voltage  and  line  current)  and  the 
resistance  losses. 

Stray  power  =  motor  input  —  resistance  losses.  (15) 

The  field  loss  may  be  eliminated  by  placing  the  ammeter  in  the 
armature  circuit  instead  of  in  the  line.  In  this  case  the  stray 


LOSSES,   EFFICIENCIES  AND   RATINGS 


99 


power  is  the  difference  between  the  armature  input  (the  product  of 
applied  voltage  and  armature  current)  and  the  losses  due  to  brush 
contact  and  armature  resistance.  (The  loss  in  the  series-field  wind- 
ing of  a  series  or  compound  dynamo  is  eliminated  by  separately  ex- 
citing the  series  field.) 


Stray  power  =  EIa  -  RJa2  -  Ia  (0.8  +  0.2 


(16) 


Resistance 


when  E   =  the  applied  electromotive  force, 

Ia  =  the  current  in  the  armature  circuit, 
Ra  =  the  resistance  of  the  armature  circuit, 
D  =  current  density  in   the  brushes   (amperes  per  square 
centimeter). 

Connections  for  the  determination  of  stray  power  in  a  shunt 
dynamo  are  shown  in  Fig.  84.  A  re- 
sistance of  the  proper  size,  considering 
the  value  of  the  current  taken  by  the 
armature,  is  connected  in  series  with 
the  armature  winding.  The  field  exci- 
tation is  regulated  by  means  of  the 
rheostat  connected  in  series  with  the 
windings,*  and  the  speed  controlled  by 
manipulating  the  resistance  in  the 
armature  circuit.  With  this  arrange- 
ment either  the  speed  or  the  field  excitation  may  be  given  any 
desired  value  within  the  operating  limits  of  the  dynamo. 

(b)  Driving  by  an  auxiliary  motor.  —  It  is  sometimes  impossible 
to  run  a  dynamo  as  a  motor  under  the  required  conditions  of  speed 

and  field  excitation,  as  when 
a  440- volt  dynamo  is  to  be 
tested  and  the  only  avail- 
able source  of  current  sup- 
ply is  no  volts.  Under 
these  conditions  a  small 
no- volt  mo  tor  is  connected 


FIG.  84.    Determination  of 
Stray  Power. 


field  Resistance 


FIG.  85.    Determination  of  Stray  Power. 


to  the  dynamo  to  be  tested,  and  the  input  to  the  auxiliary  motor 
determined  when  driving  the  dynamo  at  the  required  speed  and  with 
any  required  excitation.  Fig.  85.  The  losses  in  the  auxiliary  motor, 


*  In  the  compound  dynamo  each  winding  should  be  properly  excited. 


100 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


if  not  known,  should  be  determined  as  in  (a)  above,  and  the  stray 
power  of  the  dynamo  under  test  calculated. 

Stray  power  =  motor  input  —  motor  losses  —  resistance  losses.  (17) 

The  only  current  flowing  in  the  armature  of  the  dynamo  under 
test  is  the  small  current  required  for  the  excitation  of  its  shunt-field 
windings,  and  the  resistance  losses  in  the  armature  are  usually  so 
small  that  no  correction  need  be  made  for  them.  Series-field  wind- 
ings may  be  excited  from  the  low- voltage  mains. 

3.  Separation  of  friction  and  iron  losses.  —  It  is  evident  that  the 
friction  loss  of  a  dynamo  may  be  determined  by  driving  it,  without 
field  excitation,  by  means  of  an  auxiliary  motor.  Fig.  85. 

Frictional  losses  =  motor  input  —  motor  losses.  (18) 

Stray  power  may  be  separated  into  its  components  in  the  follow- 
ing manner:  Connect  the  dynamo  as  in  Fig.  84.  Keeping  the  speed 
constant,  determine  the  stray  power  for  different  field  excitations 


FIELD  EXCITATION  AMPERES 
FIG.  86.     Separation  of  Friction  and  Iron  Losses. 

and  plot  as  in  Fig.  86.  Extrapolate  the  curve  to  its  intersection 
with  the  axis  of  ordinates.  The  ordinate,  at  this  intersection,  repre- 
sents the  loss  due  to  friction  at  the  given  speed. 

4.   Separation  of  hysteresis  and  eddy-current  losses.  —  From 
equations  (9)  and  (12)  the  iron  loss  for  any  given  field  excitation  is 

Pt  =  kh'n  +  ke'n\  (19) 

Dividing  equation  (19)  by  n 


n 


LOSSES,  EFFICIENCIES, ^ANJT>  RAIlNGS  J>',  \  '<'.  101 


which  is  the  equation  of  a  straight  line,  and  may  be  plotted  as  in 
Fig.  87,  after  the  iron  losses  at  different  speeds  but  constant  field 
excitation  have  been  determined. 
The  hysteresis  loss  at  any  given 
or  required  speed  n  is  equal  to  the 
product   of  the   ordinate   kh'  and 


the  speed  n\  the  eddy  current  loss 
for  any  given  or  required  speed 
n  is  equal  to  the  product  of  the 
speed  and  the  ordinate  ke'n  corre- 
sponding to  the  given  or  required 
speed. 

5.   Efficiencies.  —  The  efficiency  of  an  electrical  machine  is  the 
ratio  of  the  output  to  the  input. 


SPfED 

FIG.  87.     Separation  of  Hysteresis  and 
Eddy  Current  Losses. 


Per  cent  efficiency  = 


output  X  ICQ 
input 


(21) 


The  efficiency  of  a  dynamo  is  usually  computed  from  stray  power 
and  resistance  measurements  rather  than  from  an  actual  load  test 
for  the  following  reasons: 

(a)  A  considerable  quantity  of  energy  must  be  wasted  in  making 
a  load  test. 

(b)  It  is  often  inconvenient  or  impossible  to  supply  the  electrical 
energy  required  to  run  a  large  dynamo  as  a  motor  or  to  absorb  its 
output  when  operated  as  a  generator. 

(c)  Calculated  efficiencies  are  more  reliable  than  those  determined 
by  load  test. 

The  equation  for  the  efficiency  of  a  generator,  then,  becomes 

rr    •  OUtpUt  X    ICO      .  ,       N 

per  cent  efficiency  = *-  -  »  (22) 

output  +  losses 

and  that  for  a  motor 

~  .  (input  —  losses)  100  ,    , 

per  cent  efficiency  =  * — c —  — .  (23) 

input 

The  efficiencies  of  a  dynamo  are  represented  graphically  by  means 
of  an  efficiency  curve,  per  cent  efficiency  being  used  as  ordinates 
and  per  cent  of  full  (rated)  load  as  abscissas. 

The  shunt  dynamo.  —  Very  simple  measurements  are  sufficient 
for  the  calculation  of  the  efficiency  of  a  shunt  dynamo.  The  copper 
loss  in  the  armature  is  proportional  to  the  square  of  the  current 


102  '    ESSENTIALS  ,  O.F  .ELECTRICAL  ENGINEERING 

flowing  in  the  armature  circuit  and  may  be  calculated  for  any  given 
or  required  current  after  the  resistance  of  the  armature  winding  has 
been  determined  as  explained  above. 

The  field  loss  is  approximately  constant,  the  change  in  value  due 
to  the  decrease  in  field  *  current  as  the  load  on  a  generator  increases 
is  usually  neglected,  and  is  determined  by  measuring  the  current  in 
the  field  circuit  and  the  voltage  across  its  terminals  when  the  excita- 
tion is  such  as  to  give  rated  voltage  when  operated  as  a  generator,  or 
rated  speed  when  operated  as  a  motor. 

The  stray  power  of  a  shunt  dynamo  is  approximately  constant, 
and  the  no-load  determination  is  used  in  the  calculation  of  efficien- 
cies. The  dynamo  is  operated  as  a  motor,  without  load  and  at  rated 
speed,  the  input  to  the  armature  measured,  and  the  calculated 
armature  and  brush  contact  resistance  losses  subtracted,  as  explained 
above.  The  armature  and  brush-contact  resistance  losses  are  small 
at  no-load  and  may  usually  be  disregarded  without  serious  error. 

The  fact  that  the  stray  power  of  a  shunt  dynamo  is  not  constant 
will  be  appreciated  from  the  following: 

In  a  shunt  generator  operating  at  constant  speed,  the  field  current 
falls  off  as  the  load  increases.  Therefore,  the  flux  in  the  magnetic 
circuit  is  reduced,  and  the  armature  current  changes  the  distribu- 
tion of  the  flux  in  the  air  gap.  In  the  shunt  motor,  the  speed  falls 
off  and  armature  reaction  distorts  and  reduces  the  field. 

Decreased  field  excitation,  decreased  speed,  and  armature  demag- 
netization cause  the  iron  losses  of  a  dynamo  to  decrease;  field  dis- 
tortion causes  the  iron  losses  to  increase.  These  two  effects,  there- 
fore, tend  to  neutralize  each  other.  No  accurate  and  easily  applied 
method  for  determining  the  change  in  the  stray  power  of  a  dynamo 
as  the  load  changes  has  been  devised  and,  since  it  is  usually  small, 
no  correction  need  be  attempted. 

By  differentiating  the  equation  for  the  efficiency  of  a  shunt  dy- 
namo it  may  be  shown  that  the  efficiency  is  a  maximum  when  the 
constant  losses  equal  the  variable  losses,  i.e.,  the  efficiency  is  a 
maximum  when  the  armature  resistance  losses  are  equal  to  the 
sum  of  the  field  resistance  loss  and  the  stray  power. 

The  series  dynamo.  —  Since  the  field  excitation  of  a  series  gener- 
ator, and  the  field  excitation  and  the  speed  of  a  series  motor  vary 

*  If  the  terminal  voltage  of  a  shunt  generator  remains  constant,  the  field  current 
must  be  increased  as  the  load  increases. 


LOSSES,  EFFICIENCIES   AND   RATINGS  103 

over  wide  limits,  the  stray  power  of  a  series  dynamo  also  varies 
greatly.  Therefore,  the  stray  power  must  be  determined  for  the 
particular  speed  and  field  excitation  for  which  it  is  desired  to 
calculate  the  efficiency.  By  separately  exciting  the  field  and  regu- 
lating the  voltage  applied  to  the  terminals  of  the  armature,  the  stray 
power  of  the  dynamo  for  any  speed  and  field  excitation  may  be 
determined. 

Since  the  same  current  flows  in  the  field  as  in  the  armature  of  a 
series  dynamo,  the  copper  losses  for  any  load  are  equal  to  the  square 
of  the  line  current  multiplied  by  the  combined  resistance,  in  series, 
of  the  two  windings.  The  losses  due  to  brush-contact  resistance 
should  be  calculated  as  for  a  shunt  dynamo. 

The  compound  dynamo.  —  Since  the  voltage  of  a  compound  gener- 
ator and  the  speed  of  a  compound  motor  may  vary  widely  from  their 
no-load  values,  the  efficiency  may  be  accurately  calculated  only  by 
determining  the  losses  at  the  voltage,  or  the  speed,  at  which  the 
dynamo  operates  when  carrying  the  specified  load. 

The  approximate  efficiency  of  a  compound  dynamo  obtained  by 
assuming  the  shunt-field  loss  and  the  stray  power  constant,  is  often 
sufficiently  accurate,  particularly  if  the  dynamo  is  flat  or  only 
slightly  overcompounded. 

As  a  generator.  —  Given  the  voltage  characteristic  of  a  compound 
generator,  very  close  approximations  of  the  losses  at  different  loads 
may  be  made  from  no-load  measurements.  Knowing  the  ter- 
minal voltage  at  any  condition  of  loading  and  the  resistance  of  the 
shunt-field  circuit,  the  shunt-field  loss  is  readily  calculated.  The 
resistance  losses  in  the  series  field  and  in  the  armature  are  calcu- 
lated in  the  same  way  as  for  the  series  dynamo. 

By  separately  exciting  the  series-field  winding,  the  stray  power 
of  a  compound  generator  may  be  readily  determined  for  any  re- 
quired speed  and  field  excitation.  If  the  stray  power  is  determined 
for  different  excitations,  but  constant  speed,  a  curve  may  be  plotted, 
using  stray  power  as  ordinates  and  per  cent  of  rated  load  as  ab- 
scissa. From  this  curve  the  stray  power  at  any  load  is  obtained. 

As  a  motor.  —  When  the  long  shunt  compound  dynamo  is  oper- 
ated as  a  motor  from  constant  potential  mains,  the  shunt-field  loss 
is  constant  and  the  resistance  loss  in  the  armature  and  that  in  the 
series-field  circuit  may  be  calculated  from  the  resistances  of  the 
windings  and  the  armature  current.  The  shunt-field  loss  in  a  short- 


104  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

shunt  compound  motor,  operating  from  constant  potential  mains, 
decreases  slightly  as  the  armature  current  increases  because  of  the 
drop  in  the  series-field  winding  which  reduces  the  voltage  between 
the  terminals  of  the  shunt-field  circuit.  Stray  power  should  be  de- 
termined for  speeds  corresponding  to  different  field  excitations, 
and  a  curve  plotted  as  for  a  generator. 

6.  Ratings.  —  Aside  from  the  greatly  increased  copper  losses  in 
a  dynamo  which  materially  reduce  its  efficiency  when  excessive 
current  flows  in  the  armature  windings,  the  output  is  limited  by: 
(a)  heating,  (b)  sparking,  (c)  voltage  considerations. 

(a)  Heating.  —  According  to  Joule's  Law  the  heat  liberated  in  a 
current-carrying  conductor  is  proportional  to  the  square  of  the  cur- 
rent flowing  in  the  circuit.     The  heat  due  to  hysteresis  and  eddy 
currents  also  tends  to  raise  the  temperature  of  the  armature. 

Insulations  used  on  commercial  dynamos  will  not  stand  an  in- 
definite rise  in  temperature  without  injury  to  the  insulating  quali- 
ties. The  output,  then,  must  be  kept  below  the  value  which 
causes  the  insulation  of  the  conductors  to  be  injured.  The  generally 
recognized  safe  limit  of  temperature  is  65  to  70  degrees  (Centigrade).* 
If  the  atmospheric  temperature  is  taken  as  25  degrees  (an  average 
value),  the  allowable  temperature  rise  is  40  to  45  degrees. 

(b)  Sparking.  —  Excessive  armature  currents,  by  reason  of  their 
reaction  on  the  magnetic  field,  cause  sparking  at  the  brushes  which 
burns  and  ultimately  destroys  the  commutator,  and  thus  limits  the 
output  of  a  given  dynamo. 

(c)  Voltage  limitations.  —  An  examination  of  the  magnetization 
curve  of  iron  shows  that  the  flux  in  a  given  magnetic  circuit  cannot 
be  increased  beyond  a  fixed  value  (magnetic  saturation),  without 
increasing  the  exciting  ampere-turns  beyond  the  economical  limits. 
The  alternative  is  to  increase  the  speed,  but  this  is  limited,  by 
mechanical  considerations,  to  a  peripheral  velocity  of  about  six 
thousand  feet  per  minute. 

Commutation  difficulties  increase  as  the  voltage  increases,  and 
the  commutation  of  voltages  above  1200  to  1500  is  practicable  only 
on  large  sized  machines. 

*  This  temperature  is  that  determined  by  a  thermometer,  and  is  from  5°  to  15° 
less  than  the  maximum  internal  temperature. 


LOSSES,  EFFICIENCIES   AND   RATINGS  105 

CHAPTER  VI  — PROBLEMS 

1.  A  5oo-kw.,  55<D-volt  generator  has  a  field  current  of  10  amperes,  an  arma- 
ture resistance  of  o.oi  ohm,  and  a  stray  power  of  7500  watts.     Find:  (a)  the 
efficiency  at  full  (rated)  load,  (b)  the  maximum  efficiency,  (c)  the  output  at 
maximum  efficiency. 

2.  The  dynamo  in  Problem  i  is  run  as  a  motor.     Find:  (a)  the  horse-power 
output  when  the  current  input  to  the  armature  is  750  amperes,  (b)  the  current 
input  to  the  armature  when  the  horse-power  output  is  500. 

Note.  — Output  =  El a  -  Rale?  —  S  —  Ebla. 

3.  Calculate  the  stray  power  of  a  22o-volt  shunt  motor  when  running  at  800 
r.p.m.,  the  current  input  being  15  amperes,  the  resistance  of  the  field  circuit  50 
ohms,  the  resistance  of  the  armature  circuit  0.06  ohm,  the  applied  voltage  200 
and  the  output  zero. 

4.  Calculate  the  efficiencies  of  the  motor  in  Problem  3  for  the  following  ar- 
mature currents:   25,  50,  75,  100,  125,  150,  175,  200,  225,  250,  275  and  300. 
Consider  the  field  loss  and  the  stray-power  constant. 

5.  Plot  the  efficiency  curve  from  the  values  calculated  in  Problem  4,  and 
from  the  curve  determine:  (a)  the  maximum  efficiency,  (b)  the  horse-power 
output  at  maximum  efficiency. 

6.  The  armature  resistance  of  a  25-horse-power,  230-volt  shunt  motor  is 
0.04  ohm,  and  its  stray  power  is  equal  to  5  per  cent  of  its  rated  output.     Find 
the  current  input  to  the  armature:  (a)  at  full  (rated)  load,  (b)  at  one-half  full 
load. 

7.  The  rated  current  input  to  a  23o-volt  shunt  motor  is  185.     Field  resist- 
ance =  140.    Armature  resistance  =  0.08.    No-load  speed  (with  rated  voltage 
and  normal  field  excitation)  =  785.     Current  input  at  no  load  =  8.     Find:  (a) 
stray  power,  (6)  field  loss,  (c)  armature  copper  loss  at  rated  load,  (d)  rated 
horse-power  output,  (e)  efficiency  at  full  load,  (/)  speed  regulation. 

8.  A  22o-volt  shunt  motor  has  a  stray  power  of  1000  watts  and  an  armature 
resistance  of  o.i  ohm.    At  full  load  the  armature  current  is  150  amperes  and 
the  speed  900  r.p.m.    Find:  (a)  the  no-load  speed,  (b)  the  full-load  output  in 
horse-power,  (c)  the  armature  copper  loss  at  full  load,  (d)  the  armature  current 
at  no  load. 

9.  A  23o-volt  shunt  motor  delivers  50  brake  horse  power  when  the  current 
input  is  185  amperes.     The  resistance  of  the  armature  =  0.06  ohm;  the  resist- 
ance of  the  field  circuit  =  160  ohms.     Find  the  stray  power. 

10.  A  22o-volt,  4o-horse-power  shunt  motor  has  an  armature  resistance  of 
o.i  i  ohm.     The  current  input  to  the  armature  when  operating  at  a  speed  of 
800  r.p.m.  and  delivering  the  rated  horse-power  output  is  156  amperes.    Shunt- 
field  current  =  4  amperes.     Find:   (a)  the  efficiency  at  full  load,  (b)  armature 
current  at  half  load  (2o-h.p.  output),  (c)  maximum  efficiency,  (d)  output  at 
maximum  efficiency,  (e)  speed  at  no  load. 

11.  A  shunt  dynamo,  when  operated  as  a  generator  at  800  r.p.m.,  delivers  1 1 
kw.,  the  terminal  voltage  is  no,  and  the  field  current  is  2  amperes.    Armature 
resistance  =  0.05  ohm.    When  the  same  machine  is  operated  as  a  motor,  the 


106  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

input  is  102  amperes  at  no  volts,  and  the  output  is  12.8  horse  power.    Find: 

(a)  the  counter-electromotive  force  of  the  motor,  (6)  the  speed  of  the  motor, 
(c)  the  stray  power:  (i)  of  the  generator,  (2)  of  the  motor. 

12.  A  24o-volt,  25-horse-power  shunt  motor  is  rated  to  run  at  750  r.p.m. 
Armature  resistance  =0.125  ohm.    Field  resistance  =  160  ohms.    Stray  power 
(determined  at  no  load)  =  500  watts.    Find:  (a)  speed  at  no  load,  (b)  efficiency 
at  full  load,  (c)  armature  current  at  maximum  efficiency. 

13.  The  no-load  input  to  a  22o-volt  shunt  motor  is  1.5  kw.    Armature  re- 
sistance =  0.08  ohm.     Field  resistance  =  100  ohms.     Find:    (a)   the  stray 
power,  (b)  the  field  loss,  (c)  horse-power  output  at  maximum  efficiency,  (d) 
speed  at  maximum  efficiency. 

14.  A  220-volt  shunt  motor  is  rated  at  60  horse  power.    The  maximum 
efficiency  is  90  %,  and  is  obtained  when  the  output  of  the  motor  is  50  horse 
power.    Resistance  of  the  field  circuit  =  100  ohms.    Find:  (a)  stray  power, 

(b)  resistance  of  the  armature  winding,  (c)  full  load  of  efficiency. 


CHAPTER  VII 
POLYPHASE  ALTERNATING  CURRENTS 

1.  Limitations  of  the  single-phase  system.  —  The  single-phase 
system  of  electrical  distribution  has  been  largely  displaced  for  the 
following  reasons: 

(a)  Polyphase  systems  are  more  economical  in  the  use  of  copper, 
i.e.,  a  given  power  may  be  transmitted  over  a  smaller  wire  in  the 
polyphase  system. 

(b)  Single-phase  induction  motors  are  not  self-starting  without 
auxiliary  apparatus  which  adds  to  their  mechanical  complications 
and  increases  their  cost. 

(c)  The  power  in  a  single-phase  system  is  pulsating,  that  in  a  poly- 
phase system  is  constant. 

Commercial  polyphase  systems  consist  of  two  or  more  single- 
phase  circuits  in  which  the  electromotive  forces  or  currents  are 
out  of  phase.  The  single-phase  circuits  composing  a  polyphase 
system  may  be  independent  of  each  other,  or  they  may  be  mechani- 
cally and  electrically  interconnected. 

2.  The  two-phase  system.  —  The  two-phase  system,  also  called 
four-phase  and  quarter-phase,  consists  of  two  single-phase  circuits, 
the  electromotive  forces  or  currents  of  which  are  90  degrees  out 
of  phase.    The  two  single-phase  circuits  may  be: 

(a)  independent,  (b)  interconnected  to  form  a 
three-wire  system,  (c)  star  connected,  (d)  mesh 
connected. 

(a)  Independent  phases.  —  A    two-phase    sys-  FlG-  88-   Two-phase 


tern,  the  phases  of   which  are   independent,  is  ™ 


shown  in  Fig.  88.  Evidently,  the  power  and  the 
current  in  each  phase  is  dependent  only  on  the  impedance  of  its 
load  circuit,  and  each  phase  may  be  treated  as  an  independent  single- 
phase  system. 

(b)  Three-wire  system.  —  A  three-wire  two-phase  system  is  shown, 
diagramatically,  in  Fig.  89,  one  terminal  of  each  phase  being  con- 

icy 


io8 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


nected  to  a  common  wire.  This  system  is  little  used  because  the 
voltages  between  wires  are  not  equal  and  the  current,  for  an  equal 
load  on  each  phase,  is  greater  in  the  common  wire  than  in  each  of 
the  other  two. 

If  the  voltage  between  the  terminals  of  each  phase  winding  is 
E  (  =  AB  =  BC),  the  values  and  phase  relations  are  represented 


FIG.  89.    Two-phase  Three-wire 
System. 


FIG.  90.    Vector  Diagram  of  Two-phase 
Three-wire  System 


in  a  vector  diagram,  and  the  electromotive  force  between  A  and 
C  is  the  geometric  sum  of  the  phase  voltages.     Fig.  90. 

EAC  =V(ABY  +  (BCy.  (i) 


Similarly  the  currents  in  A  and  C  may  be  represented  in  a  vector 
diagram,  and  the  current  in  B  shown  to  be  their  geometric  sum. 

IB  =  V^T+7?.  (2) 

From  the  above  it  is  evident  that  neither  the  mechanical  nor  the 
electrical  relations  are  symmetrical. 

(c)  Star  connection.  —  A  star-connected  two-phase  system  is 
shown  in  Fig.  91.  This  system  is  symmetrical,  both  mechanically 


FIG.  91.    Two-phase  Star-connected 
System. 


FIG.  92.    Two-phase  Mesh-connected 
System. 


and  electrically.     The  wires  are  all  of  the  same  size,  the  voltage  be- 

-p 

tween  alternate  wires  is  Ey  and  that  between  adjacent  wires  is  — •-• 

V2 

Each  pair  of  wires,  A  A  or  BB  in  Fig.  91,  carries  the  current  due  to  the 
load  on  that  phase  just  as  if  the  two  phases  were  not  interconnected. 
(d)  Mesh  connection.  —  The  mesh-connected  two-phase  system  is 
shown  in  Fig.  92.  The  mechanical  and  the  electrical  relations  are 
obvious  from  the  figure. 


POLYPHASE  ALTERNATING  CURRENTS  IOQ 

Two-phase  apparatus  has  not  attained  great  popularity  because 
three-phase  apparatus  offers  greater  advantages. 

3.  The  balanced  three-phase  system.  —  The  three-phase  system 
is  composed  of  three  interconnected  single-phase  circuits.  The 
three  component  circuits  may  be  connected:  (a)  star  or  wye,  (b) 
delta  or  mesh,  (c)  open  delta  or  V,  (d)  tee. 

(a)  Star  connection.  —  If  the  three  independent  circuits  shown 
in  Fig.  93  are  so  related  that  the  current  in  A  attains  its  maximum 
positive  value  one-third  of  a  cycle  before  the  current  in  B,  and  one- 


FIG.  93.     Component  Circuits  of  a  Three-     FIG.  94.     Three-phase  Star-connected 
phase  System.  System. 

third  of  a  cycle  after  the  current  in  C,  the  algebraic  sum  of  the 
currents  in  a',  b'  and  c'  is,  at  any  given  instant,  zero  and  the  ter- 
minals of  coils  A ,  B  and  C  may  be  connected  together  as  shown  in 
Fig.  94,  in  which  the  lines  a',  b'  and  c'  are  omitted.  The  maxi- 
mum values  of  current  are  assumed  to  be  equal. 

This  connection  is  known  as  star  or  wye,  and  the  junction  of  the 
three  coils  is  the  neutral  point  (or  simply  the  neutral).  A  fourth 
wire  is  sometimes  run  between  the  neutral  point  of  a  generator  and 
that  of  its  load,  but  it  carries  little  current  unless  unusual  con- 
ditions exist  in  the  system.  A  neutral  wire,  when  run,  is  always 
smaller  in  cross  section  than  are  the  mains.  A  grounding  of  the 
neutral  points  is  usually  all  that  is  required,  the  earth  acting  as  the 
neutral  conductor. 

Referring  to  Fig.  94,  let  the  electrical  relations  of  coils  A,  B  and 

C  be  such  that  •         /•     •      f  /.\ 

IA  =  1m  sin  coZ,  (3) 

IB  =  Im  sin  (ut  -  120°),  (4) 

ic  =  Im  sin  .(ut  -  240°),  (5) 

eA  =  Em  sin  co/,  ^   s  (6) 

€B  =  Em  sin  (co*  -  120°),  *  (7) 

ec  =  Em  sin  (co/  -  240°).  (8) 


110  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

Then  iA  +  iB  +  ic  =  o,  (9) 

i.e.,  the  instantaneous  currents  in  a  star-connected  system  are  posi- 
tive or  negative  as  they  flow  away  from  or  toward  the  neutral,  and 
the  sum  of  the  positive  values  is  always  equal  to  the  sum  of  the 
negative  values. 

From  equations  (6),  (7)  and  (8),  the  line-to-line  voltage  of  a  star- 
connected  system  is  SL^  A 

et  =  Em  sin  ut  —  Em  sin  (to/  =F  120°) ;  (10) 

the  voltage  between  the  line  terminals  of  coils  A  and  B  is  maxi- 
mum when  co/  =Jto°  or  240°;  the  voltage  between  the  line  termi- 
nals of  coils  A  and  C  is  maximum  when  co/^=  120°  or  300°;  and 
the  voltage  between  the  line  terminals  of  coils  B  and  C  is  maxi- 
mum when  co/  =  180°  or  o°. 
Therefore, 

(E,)m  =  (Ep)n  sin  60°  -  (Ep}m  sin  300°.  (11) 

=  0.866  (Ep)m  +  0.866  (Ep)m  (12) 

=  1.732  (Ep)m  (13) 

=  ^3  (Ep)m               ,  (14) 

and                         EZ  =  V3£P.  (15) 

i.e.,  the  effective  line-to-line  voltage  of  a  balanced  star-connected 
system  is  equal  to  the  effective  phase  voltage  multiplied  by  the 
square  root  of  3. 

It  is  evident,  from  the  connection,  that  the  same  current  flows  in 
the  coil  of  a  star-connected  system  as  flows  in  the  line  to  which  the 
coil  is  connected. 

Since  the  line  current  and  the  phase  electromotive  force  of  a 
balanced  star-connected  system  supplying  a  non-reactive  load  cir- 


FIG.  95.    Vector  Diagram  of  Star-con-  FIG.  96.    Three-phase  Delta-con- 

nected System.  nected  System. 

cuit  attain  their  maximum  values  at  the  same  instant,  it  follows 
that  the  line  current  and  the  line-to-line  voltage  are  out  of  phase 
by  30  degrees,  as  shown  in  Fig.  95. 


POLYPHASE  ALTERNATING  CURRENTS  III 

(6)  Delta  connection.  —  Referring  to  Fig.  96,  let  the  electrical  re- 
lations of  coils  A ,  B  and  C  be  such  that 

eA  =  Em  sin  w/,  (16) 

eB  =  Em  sin  (co*  -  120°),  (17) 

ec  =  Em  sin  («J  —240°),  (18) 

*  A  =  /m  sin  w/,  (19) 

/ 

iB  =  /m  sin  («/  —  120°),  (20) 

*c  =  I™  sin  (w/  —  240°),  (21) 

Then  eA  +  eB  +  ^c  =  o,  (22) 

f.e.,  the  instantaneous  electromotive  forces  in  a  delta-connected 

system  are  negative  or  positive  as  they  tend  to  cause  a  current  to 

flow  in  a  clockwise  or  in  a  counter-clockwise 

direction  around  the  circuit  formed  by  the  three 

coils,  and  the  sum  of  the  positive  electromotive 

forces  is  always  equal  to  the  sum  of  the  negative 

electromotive  forces. 

In  a  delta-connected  system  the  voltage  be- 
tween lines  is  the  voltage  between  the  terminals  FlG-  97*    Vector  Dia~ 
r  ,1  .11  ,1  'if,!  M  gram  of  Delta-con- 

of  the  coil,  because  the  terminals  of  the  coil  are      nected  System 

connected  directly  to  the  lines. 

From  equations  (19),  (20)  and  (21),  the  line  current  in  a  delta- 
connected  system  is 

ii  =  Im  sin  wt  -  Im  sin  («/  =F  120°) ;  (23) 

the  current  in  the  line  to  which  coils  A  and  B  are  connected  is 
maximum  when  co/  =  _6o^orj24°° >  the  current  in  the  line  to  which 
coils  A  and  C  are  connected  is  maximum  when  co/  =  120°  or  300°,* 
and  the  current  in  the  line  to  which  coils  B  and  C  are  connected 
is  maximum  when  co/  =  180°  or  o°. 
Therefore, 

(/i)«  =  (/„)„  sin  60°  -  (IP)m  sin  300°  (24) 

=  0.866  (/p)m  +  0.866  (IP)m         (25) 
=  1.732  (IP)m  (26) 

=  V3(U«  (27) 

and  /,  =>/3  /,.  (28) 


112  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

i.e.,  the  effective  line  current  in  a  balanced  delta-connected  system 
is  equal  to  the  effective  current  in  each  winding  multiplied  by  the 
square  root  of  3. 

Since  the  line-to-line  voltage  and  the  phase  current  in  a  balanced 
delta-connected  system  supplying  a  non-reactive  load  circuit  attain 
their  maximum  values  at  the  same  instant,  it  follows  that  the  line 
current  and  the  line-to-line  voltage  are  30  degrees  out  of  phase. 

(c)  V -connection.  —  With  the  open-delta  or  V-connection,  a  three- 
phase  system  is  obtained  by  the  use  of  only  two  windings.  Fig.  98. 
The  connection  is  essentially  that  of  the  delta  with  one  coil  omitted. 


FIG.  98.    Three-phase  V-connection.      FIG.  99.    Vector  Diagram  of  V-connection . 

Obviously,  the  line-to-line  voltage  is  that  of  the  winding,  and  the 
line  current  must  flow  in  the  winding.  The  vector  diagram  for  a 
V-connection  is  shown  in  Fig.  99. 

Let  the  load  on  a  delta-connected  system  be  300  kw.  at  unity 
power  factor.     It  is  obvious  that  each  winding  carries  one-third  of 
the  total  load  or  ioo  kw.     If  the  same  load  is  carried  by  an  open- 
delta  system,  each  winding  must  carry  one-half  the  total  or  150  kw., 
but  the  current  and  the  electromotive  force  in 

c       

the  windings  are  30  degrees  out  of  phase  and  the 
current-carrying  capacity  (rating)  of  each  open- 
"    delta  winding  must  be  1.73  times  the  current- 
carrying  capacity  of  each  delta  winding.     The 
'  weight  of  copper  required  for  an  open-delta  con- 
FIG.  ioo.    Three-phase  nection  is,  therefore,  15  per  cent  greater  than 

that  required  for  a  delta  connection. 

(d)  T-connection.  —  A  three-phase  system  is  obtained  if  two  coils, 
the  electromotive  forces  in  which  are  in  quadrature,  are  connected 
as  indicated  in  Fig.  ioo. 
Referring  to  Fig.  ioo,  let 

EAB  =  EBC  =  ECA  (29) 

EEC  \ 

and  ECD  ==  EBD  =  ~~  —  *  \3w 

2 


POLYPHASE  ALTERNATING   CURRENTS  113 

The  voltages  EAB,  EBD  and  EAD  are,  by  construction,  the  hypothe- 
nuse,  the  base  and  the  altitude  of  a  right  triangle.  Fig.  101. 
Therefore,  _ 

(31) 


=  0.866  EAB,  (32) 

i.e.,  a  three-phase  system  having  equal  voltages  between  lines  is 
produced  by  two  windings  connected  as  in  Fig.  100,  if  the  elec- 
tromotive forces  in  the  windings  are  in  quadrature  and  that  in  AD 
is  0.866  times  that  in  BC. 
Let  the  current  in  winding  AD  be  in  phase  with  the  voltage 

EAD.     Then 

*A  =  /msinco/,  (33) 

is  =  Im  sin  («/  -  120°),  (34) 

.   m  ic  =  Im  sin  («/  -  240°).  (35) 

But  eBC  =  (EBc)m  sin  (co/  -  90°).  (36) 

Therefore,  at  unity  power  factor,  the  current  in  one-half  of  the  wind- 

ing BC  leads  the  electromotive  force,  and  that  in 

the  other  half  lags  behind  the  electromotive  force, 

as  shown  in  Fig.  101.  ,/ 

4.  Comparison  of  star  and  delta  connec- 
tions. —  The  star-connected  three-phase  system  FIG.IOI.  Vector  Dia- 
has  the  following  advantages  over  the  delta-  gram  of  T-conneo 
connected  system: 

(a)  The  availability  of  a  neutral  point  to  which  a  ground  wire, 
a  meter  connection  or  a  load  may  be  connected. 

(b)  Circulating  currents  cannot  flow  in  the  windings  of  a  star- 
connected  system. 

(c)  For  a  given  line-to-line  voltage,  the  number  of  turns  required 
in  a  star-connected  armature  is  only  58  per  cent  of  the  number 
required  in  a  delta-connected  armature. 

(d)  The  electromotive-force  wave  of  a  star-connected  armature 
winding  is  more  nearly  harmonic  than  that  of  a  delta-connected 
winding.* 

*  A  non-harmonic  electromotive  force  wave  may  be  resolved  into  a  fundamental 
sine  wave  and  the  odd  harmonics  of  this  fundamental.  By  reason  of  their  phase 
relations,  some  of  these  harmonics,  particularly  the  third,  cause  currents  to  flow  in 
the  circuit  formed  by  delta-connected  coils,  but  neutralize  each  other  in  star-  connected 
coils  and  do  not  appear  in  the  line-to-line  voltage  wave. 


114  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

5.  Unbalanced  three-phase  system.  —  When  the  load  on  a  three- 
phase  system  is  unequally  divided  between  the  three  phases,  the 
system  is  said  to  be  unbalanced,  and  the  currents  flowing  in  the 
different  lines  are  no  longer  equal.  The  voltages  between  lines  are 
also  unbalanced  to  a  slight  degree.  When  the  line  currents  are 
unequal,  the  algebraic  sum  of  the  instantaneous  line  currents  may 
not  be  zero  and  an  equalizing  current,  IN  in  Fig.  102,  flows  in  the 
neutral  of  a  star-connected  system,  or  distorts  the  current  triangle 


FIG.  102.    Vector  Diagram  of  Unbalanced   FIG.  103.    Vector  Diagram  of  Unbalanced 
Star-connected  System.  Delta-connected  System 

of  a  delta-connected  system,  as  indicated  in  Fig.  103.  If  the  star- 
connected  system  is  not  provided  with  a  neutral  connection,  the 
current  relations  are  still  further  distorted. 

6.  Power  factor  of  a  polyphase  system.  —  The  term  "  power 
factor,"  when  applied  to  a  polyphase  system,  can  have  no  rational 
meaning  since  each  component  circuit  may  have  a  different  power 
factor. 

7.  Power  in  a  polyphase  system.  —  It  has  been  shown*  that  the 
power  in  a  single-phase  alternating-current  circuit  is  equal  to  the 
product  of  the  electromotive  force,  the  current  and  the  cosine  of 
the  angle  of  phase  difference. 

P1  =  £p/pcos0.  (37) 

The  power  in  a  balanced  two-phase  system  is,  therefore,  twice  that 
in  one  of  the  component  circuits, 

P2  =  2£p/pCOS0,  (38) 

and  that  in  a  balanced  three-phase  system  is  three  times  that  in 
one  of  the  component  circuits. 

Wz  =  3  EPIP  cos  <£,  (39) 

when  Ep  =  the  phase  voltage, 

IP  =  the  phase  current, 
cos  0  =  the  power  factor  of  each  component  circuit. 

*  See  Chapter  i,  Section  28. 


POLYPHASE  ALTERNATING   CURRENTS  115 

The  line  quantities  of  a  polyphase  system  are  often  more  easily 
measured  than  the  phase  quantities,  and  it  is  convenient  to  have  a 
power  equation  into  which  the  line  quantities  may  be  substituted. 
From  equation  (15),  the  line-to-line  voltage  of  a  star-connected 
system  is  equal  to  A/3  times  the  phase  voltage;  from  equation  (28), 
the  line  current  in  a  delta-connected  system  is  A/3  times  the  phase 
current.  Therefore, 

P3  =  V3 -Ei/i  cos  0,  (40) 

when  EI  =  the  line-to-line  voltage, 

1 1  =  the  line  current, 
cos  c/>  =  the  power  factor  of  each  component  circuit. 

8.  Constancy  of  power  in  a  balanced  polyphase  system.  —  One 
of  the  advantages  of  the  polyphase  system,  when  compared  with  the 
single-phase  system,  is  its  constancy  of  power,  i.e.,  the  torque  of  a 
single-phase  motor  is  pulsating;  that  of  a  polyphase  motor  is 
constant. 

Let  the  electromotive  force  and  current  relations  in  one  phase  of  a 
two-phase  system  be  such  that 

ef  =  Em  sin  co/,  (41) 

*'  =  Im  sin  (co/  -  0).  (42) 

Then  the  relations  in  the  second  phase  are 

e"  =  Em  sin  (co/  -  90°)  (43) 

=  Em  cos  co/,  (44) 

i"  =  Im  sin  («*  -  90°  -  </>)  (45) 

=  Im  cos  (co/  -  0),  (46) 

and        p2  =  e'i'  +  e"i"  (47) 

=  Emlm  [sin  co/  sin  (co/  —  0)  +  cos  co/  cos  (co/  —  0)].       (48) 

Expanding  equation  (48) 

p2  =  Emlm  cos  0  (sin2  co/  +  cos2  co/).  (49) 

But  sin2  co/  +  cos2  co/  =  i.  (50) 

Therefore,  />2  =  £m/m  cos  t/>,  (51) 

which  is  a  constant. 

In  a  similar  manner  it  may  be  shown  that  the  power  in  a  three- 
phase  system  is  constant,  and  equal  to 

pa  =  1.5  Emlm  cos  <t>.  (52) 


n6 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


9.  Power  measurements.  —  The  power  in  a  single-phase  system 
may  be  measured  by  means  of  a  wattmeter  connected  as  indicated 
in  Fig.  104.  Since  a  polyphase  system  is  composed  of  two  or 

more  single-phase  circuits,  the  power 
in  a  polyphase  system  is  the  sum  of 
the  indications  of  wattmeters  con- 
nected in  the  component  single-phase 
circuits. 


.-Current  Coil 


Voltage  Coil 


FIG.  104.    Single-phase. 


In  the  two-phase  system  a  wattmeter  connected  in  either  phase 
indicates  half  the  total  power,  if  the  system  is  balanced;  if  the 


Wattmeter 


FIG.  iosa.    Two-phase  Four-wire. 


FIG.  io5b.    Two-phase  Three-wire. 


Co) 


system  is  unbalanced,  a  wattmeter  must  be  connected  in  each  phase 
and  the  sum  of  the  readings  taken  as  the  total  power  of  the  system. 

The  connections  for  a  four-wire  sys- 
tem, either  independent  or  intercon- 
nected, are  shown  in  Fig.  io5a,  and 
those  for  a  three-wire  system  in 
Fig.  io5b. 

The  power  in  a  balanced  three- 
phase  system  may  be  determined 
from  the  indication  of  one  wattmeter 
connected  as  shown  in  Fig.  io6a. 
The  current  coil  of  the  wattmeter  is 
connected  into  any  one  of  the  three 
lines,  and  the  potential  coil  is  con- 
nected between  this  line  and  the  neu- 
tral of  a  star-connected  system.  In 
a  delta-connected  system,  or  in  a  star- 
connected  system  the  neutral  of  which 


Resistance 


Cb) 


•Waff-meter 


Wattmeter 


FIG.  106.    Three-phase. 


is  inaccessible,  a  so-called  "artificial  neutral"  may  be  constructed 
by  using  two  resistances,  each  of  which  is  equivalent  to  the  resist- 
ance of  the  potential  circuit  of  the  wattmeter.  When  connected 
as  shown  in  Fig.  io6b,  the  resistances  and  the  potential  circuit  of 


POLYPHASE  ALTERNATING  CURRENTS  117 

the  meter  form  a  star  connection,  and  the  meter  indicates,  for 
balanced  load,  one-third  the  total  power  of  the  system. 

A  more  generally  used  method  for  determining  the  power  in  a 
three-phase  circuit  makes  use  of  two  wattmeters  connected  as  indi- 
cated in  Fig.  io6c.  The  algebraic  sum  of  the  indications  of  the 
meters  is  the  total  power  in  the  system,  whether  the  system  is 
balanced  or  unbalanced.*  The  current  coils  of  the  meters  are  con- 
nected into  any  two  of  the  three  lines,  and  the  potential  coils  be- 
tween these  two  lines  and  the  third  line.  The  wattmeter  indica- 
tions are  equal  only  when  the  power  factors  of  the  load  circuits 
are  unity  and  the  load  is  balanced. 

In  using  two  wattmeters  for  measuring  the  power  of  a  three- pliase 
system,  the  sum  of  their  indications  is  to  be  taken  if  the  power  fac- 
tors of  the  load  circuits  are  greater  than  50  per  cent  and  their  dif- 
ference if  the  power  factors  are  less  than  50  per  cent.  The  power 
factors  of  induction  motors  running  at  light  loads  are  often  very 
low.  To  determine  whether  the  power  factor  of  a  circuit  is  greater 
or  less  than  50  per  cent,  disconnect  the  potential  coil  of  wattmeter 
No.  i  from  line  B  and  connect  it  to  line  A.  Fig.  io6c.  If  the 
direction  of  torque  (the  direction  in  which  the 
pointer  deflects)  is  not  reversed,  the  power  factor 
is  greater  than  50  per  cent. 

Referring  to  Fig.  107,  let  OA,  OB  and  OC 
represent  the  equal  line  to  neutral  voltages  of 
a  balanced  three-phase  system,!  the  power  factor 
of  which  is  50  per  cent  (angle  of  lag  equals  60°). 
The  line  currents,  then,  are  represented  by  the  FIG.  107.  Vector  Dia- 
vectors  IA,  IB  and  Ic  and  the  line  voltages  by       Deters 
AB,  BC  and  AC. 

If  the  currents  in  the  coils  of  a  wattmeter  remain  constant,  the 
indication  of  the  meter  is  proportional  to  the  cosine  of  the  angle  of 
their  phase  difference.  Therefore,  wattmeter  No.  2,  with  a  phase 
difference  of  90  degrees,  indicates  zero,  and  wattmeter  No.  i,  with 
a  phase  difference  of  30  degrees,  indicates  the  total  load  in  the 
three-phase  system.  If  the  power  factor  of  the  system  is  greater 
than  50  per  cent,  the  indication  of  wattmeter  No.  2  is  positive  and 

*  An  unbalanced  three-phase  system  with  "grounded"  neutral  is  essentially  a  four- 
wire  system,  and  three  wattmeters  should  be  used  if  accurate  determination  of  the 
power  is  required. 

t  It  is  immaterial  whether  the  system  is  star-  or  delta-connected. 


Il8  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

must  be  added  to  the  indication  of  wattmeter  No.  i  to  obtain  the 
total  power  in  the  system;  if  the  power  factor  of  the  system  is  less 
than  50  per  cent,  the  direction  of  current  in  one  coil  of  wattmeter 
No.  2  must  be  reversed  to  bring  the  indication  on  the  scale,  and 
its  indication  regarded  as  negative.  . 

From  the  above  considerations  it  is  evident  that 

PI  =  EABIA  cos  (30  -  0)  (53) 

and  P2  =  EBcIc  cos  (30  +  0),   .  (54) 

when     PI  =  the  indication  of  wattmeter  No.  i, 
PI  =  the  indication  of  wattmeter  No.  2, 
EAB  =  EEC  =  the  line-to-line  voltage, 
I  A  —  Ic  =  the  line  currents, 
0  =  the  power-factor  angles. 

It  was  shown  in  Section  7  that  the  total  power  in  a  balanced 
three-phase  system  is 

P3  =  VsEj/jcos*.  (55) 
But            Pi  +  P2  =  Elll  [cos  (30°  -  0)  +  cos  (30°  +  0)  ]         (56) 
=  EJi  (cos  30°  sin  0  +  sin  30°  sin  <f> 

4-  cos  30  sin  <j>  -  sin  30°  sin  <£)  (57) 

=  2  Eil  i  cos  30°  cos  0  (58) 

=  A/3  EJ  i  cos  0.  (59) 

Therefore,  the  total  power  of  a  three-phase  system  is  the  algebraic 
sum  of  the  indications  of  two  wattmeters  connected  as  shown  in  Fig. 
io6c,  and  the  indications  of  the  meters  are  equal  only  when  the 
phase  angle  $  is  unity,  i.e.,  when  the  load  circuits  are  non-reactive. 
The  power  factor  of  a  balanced  three-phase  system  may  be  cal- 
culated from  the  wattmeter  readings  and  the  following  formula: 

COS  0    =  *  r  ,  (60) 


when      n  =  -=• 


PI  =  the  indication  of   that  wattmeter  which   is  always 

positive, 
Pz  =  the  indication  of  that  wattmeter  which  may  be  either 

positive  or  negative. 


POLYPHASE  ALTERNATING  CURRENTS  1  19 

10.  Comparison  of  two-  and  three-phase  systems.  —  The  state- 
ment was  made  in  Section  2  that  the  three-phase  system  is  superior 
to  the  two-phase  system.  This  superiority  lies  chiefly  in  the  smaller 
weight  of  copper  required  in  the  three-phase  system  for  the  trans- 
mission or  the  distribution  of  a  given  power,  the  line  voltage  and 
the  power  lost  in  the  two  systems  being  equal.  Let 

E  =  the  greatest  line-to-line  voltage  in  each  system, 

72  =  the  current  in  each  line  of  the  two-phase  system, 

73  =  the  current  in  each  line  of  the  three-phase  system, 

R2  =  the  resistance  of  each  of  the  four  conductors  in  the  two- 

phase  system, 
R3  —  the  resistance  of  each  of  the  three  conductors  in  the 

three-phase  system. 

Then,  2EI2  =  V^EIS  (61) 


and  72  =  .  (62) 

The  copper  loss  in  the  two-phase  system  is 

P2  =  4#2/22  (63) 

and  that  in  the  three-phase  system  is 

PS  =  3  *3/32.  (64) 

Since  the  copper  losses  in  the  two  systems  are  equal 

4#2/22   =  3^3/32.  (65) 

Substituting  the  value  of  72  from  equation  (62) 

3*3/32  (66) 


4 
and  R2  =  R3,  (67) 

i.e.,  the  resistance  of  each  conductor  in  the  two-phase  system  is 
equal  to  the  resistance  of  each  conductor  in  the  three-phase  system, 
and  the  total  weight  of  conductor  in  the  two-phase  system  is  one- 
third  greater  than  that  in  the  three-phase  system. 

When  the  three-wire  two-phase  system  having  a  phase  voltage 
equal  to  the  line-to-line  voltage  of  a  three-phase  system,  is  compared 
with  the  three-phase  system,  the  weight  of  copper  in  the  two-phase 
system  is  found  to  be  2.8  per  cent  less  than  that  required  in  the 


120  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

three-phase  system.  This  small  saving  in  copper  is  more  than 
offset  by  the  increased  insulation  required  by  reason  of  the  40  per 
cent  larger  voltage  between  two  of  the  conductors. 

n.  Equivalent  single-phase  system.  —  In  problems  involving 
polyphase  quantities,  it  is  often  convenient  to  consider  the  poly- 
phase system  replaced  by  an  equivalent  single-phase  system,  i.e., 
by  a  single-phase  system,  the  voltage  of  which  is  equal  to  the 
greatest  voltage  between  lines  in  the  polyphase  system,  and  in 
which  the  total  power,  the  losses  and  the  power  factor  are  equal 
to  those  in  the  polyphase  system. 

Equivalent  of  the  two-phase  system.  —  Since  the  two-phase  system  is 
composed  of  two  single-phase  circuits,  the  voltage  of  the  equivalent 
single-phase  system  is  the  phase  voltage  of  the  two-phase  system,* 
and  the  equivalent  single-phase  current  is  twice  that  in  each  of  the 
two-phase  conductors.  The  copper  loss  in  the  two-phase  system  is 

P2  =  2£2/22,  (68) 

when  jR2  =  the  resistance  of  each  phase  circuit, 
/2  =  the  current  in  each  line. 

The  copper  loss  in  the  equivalent  single-phase  circuit  is 

P1=Rl(2l2)2.  (69) 

But  Ri  (2  72)2  =  2  £2/22  (70) 

and  *i  =  ^>  (71) 

2 

i.e.,  the  resistance  of  the  equivalent  single-phase  system  is  equal  to 
one-half  the  resistance  of  each  circuit  of  the  two-phase  system. 

Equivalent  of  the  three-phase  system. — Assuming  unity  power 
factor,  the  power  in  a  three-phase  system  is 

P3  =  V3£/3  (72) 
and  that  in  the  equivalent  single-phase  system  is 

Pi  =  Eh.       -  (73) 

Therefore,  -  /i  =  V3  73.  (74) 

tfi/i2  =  3*3/32  (75) 

and  #1  =  -Rs,  (76) 

*  Unless  the  phases  are  interconnected  to  form  a  three-wire  system. 


POLYPHASE  ALTERNATING   CURRENTS  121 

when  RI  =  the  resistance  of  the  equivalent  single-phase  system, 
Rs  =  the  resistance  of  each  phase  of  the  three-phase  system, 
/i  =  the  current  in  the  equivalent  system, 
73  =  the  current  in  each  line  of  the  three-phase  system, 

i.e.,  the  resistance  of  the  equivalent  single-phase  system  is  equal 
to  the  resistance  of  each  phase  of  the  three-phase  system. 

It  is  evident  from  inspection  that  the  equivalent  resistance  of  a 
star-connected  system  is  one-half  the  resistance  as  measured  by  con- 
tinuous-current methods  between  any  two  lines  of  the  three-phase 
system.  It  may  be  shown  that  this  is  also  true  for  the  delta-con- 
nected system. 

CHAPTER  VII  — PROBLEMS 

1.  A  balanced  3-wire,  2-phase  system  delivers  50  kw.  at"  a  voltage  (phase) 
of  220  and  a  power  factor  of  0.9.     Find:  (a)  the  current  in  each  line  wire,  (b)  the 
voltage  between  the  "outside"  wires. 

2.  The  current  in  the  common  wire  of  Problem  i  flows  in  the  current  coil 
of  a  wattmeter,  and  the  voltage  coil  of  the  wattmeter  is  connected  between  the 
common  wire  and  one  of  the  "outside"  wires.    Find  the  wattmeter  indica- 
tion. 

3.  The  current  in  the  common  wire  of  Problem  i  flows  in  the  current  coil  of  a 
wattmeter  and  the  voltage  coil  of  the  wattmeter  is  connected  between  the  "out- 
side" wires.     Find  the  wattmeter  indication. 

4.  A  3-phase,  star-connected  system  has  a  line-to-line  voltage  of  6600. 
Find  the  phase  voltage  (line  to  neutral). 

5.  The  line  current  in  a  balanced  delta-connected  system  is  150.    Find  the 
phase  current. 

6.  Find  the  power  in  the  circuit  of  Problem  5  when  the  phase  voltage  is  240, 
and  the  power  factor  of  the  load  circuit  is  0.85. 

7.  The  current  coil  of  a  wattmeter  is  connected  in  one  line  of  a  balanced  3- 
phase  system,  and  the  voltage  coil  of  the  wattmeter  between  the  other  two  lines. 
The  kw.  output,  of  the  system  is  100,  the  voltage  220  and  the  power  factor 
unity.     Find  the  indication  of  the  wattmeter. 

8.  A  non-inductive  star-connected  load  and  a  non-inductive  delta-connected 
load  are  operated  from  the  same  3-phase  system.    The  phase  current  in  each 
load  circuit  is  100  amperes.    Find  the  current  in  each  line  of  the  supply  system. 

9.  Same  as  Problem  8  except  the  power  factor  of  the  star-connected  loads 
is  0.866.     Find  the  line  currents. 

10.  The  indications  of  two  wattmeters  connected  as  in  Fig.  io6c  are  5000  and 
2500.    The  system  is  balanced.    Find  the  power  factor  of  the  load  circuits. 

11.  The  phase  currents  in  a  delta-connected  3-phase  system  are  50,  30  and 
75  amperes,  the  load  circuits  being  non-inductive.    Find  the  current  in  each 
line.     Solve  graphically. 


122  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

12.  The  phase  currents  of  a  2-phase,  3-wire  system  equal  160  amperes.    The 
power  factor  of  the  load  connected  to  one  phase  is  0.85,  that  of  the  load  con- 
nected to  the  other  phase  is  unity.     Find  the  current  in  the  common  wire. 

13.  Determine  the  single-phase  equivalent  of  a  balanced  3-phase  system 
delivering  500  kw.  at  2300  volts,  and  having  a  power  factor  of  0.92. 

14.  The  line  current  in  a  mesh-connected  2-phase  system  (balanced)  is  100 
amperes.     Find  the  current  in  each  coil  (Fig.  92). 

15.  Find  the  power  in  the  2-phase  system  in  Problem  14  if  the  voltage  be- 
tween alternate  wires  (Fig.  92)  is  2300. 

16.  Three  resistances,  the  values  of  which  are  10,  15  and  20  ohms,  are  star- 
connected  to  a  3-phase  system,  the  line-to-line  voltage  of  which  is  220.     Find: 
(a)  the  current  in  each  line,  (b)  the  voltage  between  each  line  and  neutral,  (c) 
the  power  input  to  the  resistances. 

17.  100  kw.  are  delivered  to  a  balanced  3-phase  system,  the  power  factor  of 
which  is  0.85,  and  the  voltage  between  lines  is  2200.    Find  the  current  flowing 
in  each  line. 


CHAPTER  VIII 
THE  ALTERNATING-CURRENT   GENERATOR 

i.  Voltage.  —  The  voltage  of  an  alternating-current  generator 
is  affected  by  the  same  quantities  that  affect  the  voltage  of  a  con- 
tinuous-current generator,  and  in  addition,  by  the  relative  posi- 
tions of  the  armature  conductors.  The  equation  for  the  voltage 
of  an  alternating-current  generator  may  be  written 

Ea  =  k<f>npN  io-8,  (i) 

when  Ea  =  the  effective  electromotive  force  induced  in  the  arma- 
ture winding, 

k  =  a  constant,  the  value  of  which  depends  on  the  rela- 
tive positions  of  the  armature  conductors, 
<j>  =  the  total  flux  passing  between  each  pole  and  the  arma- 
ture, 


(a)  Stator  (armature)  (b)  Field  Structure 

FIG.  108.    Revolving  Field  Alternator.    Triumph  Electric  Co. 

n  —  the  speed  of  the  armature  or  of  the  rotating  field 

in  revolutions  per  second, 
p  =  the  number  of  poles  in  the  field  structure, 
N  =  the  number  of  series  conductors  on  the  surface  of  the 
armature. 

123 


124  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

Since  the  product  of  the  speed  and  the  number  of  field  poles  of 
an  alternator  is  equal  to  twice  the  frequency,  equation  (i)  may  be 

written 

Ea  =  2  jkfN  io-8,  (2) 

which  is  often  a  useful  form  of  the  expression. 

2.  Concentrated  armature  windings.  —  Concentrated  armature 
windings  are  those  in  which  all  the  conductors  under  any  given 
pole  are  placed  in  one  slot,  as  indicated  in  Fig.  ioga.  It  is  evident 
that  the  same  electromotive  force  is  induced  in  each  of  the  con- 

ductors, and  that  the  maximum  values 
are  attained  at  the  same  instant. 
Therefore,  the  total  electromotive  force 
induced  in  the  winding  is  that  in- 
duced in  each  conductor  multiplied 

FIG.  iooa.     Concentrated  Anna-     ,       .  ,  i          e          ^  ^ 

ture  Winding  PX  tne  number  of  conductors  connected 

in  series. 

Since  the  quantity  <f>npN  io~8  is  the  average  electromotive 
force  induced  in  the  armature  winding,  the  value  of  k  for  a  con- 
centrated winding  is  the  ratio  of  the  effective  to  the  average  value. 
For  a  sine  wave 


=  I-  II-  (4) 

3.  Distributed  armature  windings.  —  Distributed  armature 
windings  are  those  in  which  the  conductors  under  a  given  pole 
are  divided  into  groups,  and  each  group  is  placed  in  a  different  slot, 
as  indicated  in  Fig.  logb.  With  this  arrangement  of  the  armature 
conductors,  the  electromotive  force  induced 
in  each  conductor  is  the  same,  but  the  maxi- 
mum values  are  not  attained  at  the  same 
instant  in  the  different  groups,  i.e.,  the 
electromotive  force  induced  in  each  group 
is  out  of  phase  with  the  electromotive  FlG,  109b'  £!st,ributed 

Armature  Winding. 

forces  induced  in  the  other  groups.     There- 

fore, the  electromotive  force  of  an  alternating-current  generator, 
having  a  distributed  armature  winding,  is  the  geometric  sum  of  the 
electromotive  forces  induced  in  the  several  groups  into  which  the 
armature  winding  is  divided. 


THE  ALTERNATING-CURRENT  GENERATOR 


125 


ii  the  armature  winding  consists  of  two  similar  groups  of  conduc- 
tors per  pole  as  in  Fig.  1090,  the  maximum  in  group  B  occurs  one- 
quarter  of  a  cycle  (90  electrical  degrees)  later  than  the  maximum  in 
group  A .  The  electromotive  forces  of  the  two  groups  are,  therefore, 
90  degrees  out  of  phase,  and  the  resultant  electromotive  force  is 
\/2  times  that  induced  in  one  group.  For  any  number  of  groups 
into  which  the  armature  conductors  may  be 
divided,  the  values  and  the  phase  relations 
of  the  electromotive  forces  in  the  different 
groups  may  be  plotted,  and  the  resultant 
electromotive  force  found  graphically.  Fig. 
no.  In  calculating  the  effective  electro- 
motive force  induced  in  the  armature  of  an 
alternating-current  generator,  it  should  be 
remembered  that  each  group  is  a  concen- 
trated winding,  and  that  the  effective  elec- 
tromotive force  induced  in  it  is  the  average 
electromotive  force  multiplied  by  I.H,  as  explained  for  the  concen- 
trated winding. 

From  Fig.  no  it  is  evident  that 


FIG.  no.    Vector  Diagram 
of  Distributed  Winding. . 


k  = 


i.nOd 


•f  Ob  +  Oc  +  Od 

1. 1 1  (chord  of  the  group  arc) 
sum  of  chords  between  groups 


(s) 

(6) 


For  an  infinite  number  of  groupings  of  the  conductors,  the  sum  of 
the  chords  becomes  an  arc,  as  indicated  by  the  dotted  arc  in  Fig. 

no. 

TABLE  III 
VALUES  OF  k  FOR  DISTRIBUTED  ARMATURE  WINDINGS 


No.  of  slots 


Degrees  subtended  by  the  winding 


per  pole 

180 

135 

90 

60 

45 

2 

0.784 

0.922 

1.025 

1.072 

1.087 

3 

o-739 

0.893 

1.  012 

1.065 

1.084 

4 

0.724 

0.882 

1.007 

1.063 

1.083 

Infinite 

0.707 

0.876 

I  .OOO 

I  .060 

1.082 

126  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

4.  Effects  of  distributed  armature  windings.  —  The  effects  of 
distributing  the  armature  winding  of  a  alternator  are:  (a)  to  de- 
crease the  induced  electromotive  force,  (b)  to  reduce  the  induc- 
tance* of  the  armature  circuit  and  improve  the  regulation,  (c)  to 
make  the  induced  electromotive  force  more  nearly  harmonic,  (d)  to 
distribute  the  heating  over  the  armature  surface,  (e)  to  reduce  the 
size  of  the  armature. 

The  effect  of  a  distributed  winding  on  the  shape  of  the  electro- 
motive-force wave  is  clearly  shown  in  Fig.  in.    The  electromotive- 

A    A    A    A  A    A    A   / 

v  v  v          /  V  V  V 

FIG.  ma.    Voltage  Wave  with  Concen-       FIG.  nib.    Voltage  Wave~with  Dis- 
trated  Armature  Winding.  tributed  Armature  Winding. 

force  wave  of  the  concentrated  winding  (Fig.  ma)  shows  a  very 
pronounced  third  harmonic  which  is  entirely  absent  from  the  wave 
of  the  distributed  winding  (Fig.  nib).  The  harmonics  of  the 
different  groups  of  a  distributed  winding  tend  to  neutralize  and 
eliminate  the  harmonic  from  the  resultant  electromotive-force 
wave. 

5.  The  oscillograph.  —  The  shape  of  an  alternating  current  or 
electromotive-force  wave  is  determined  by  means  of  the  oscillo- 
graph.    A  very  fine  wire  or  strip  is  bent 
into  a  loop  and  placed  between  the  poles 
of  a  powerful  electromagnet,  as  indicated  in 
Fig.  112.    Attached  to  the  loop  is  a  small 
mirror  M  on  which  may  be  focused  a  beam 
of  light. 

When  current  flows  in  the  looped  conduc- 
FIG-  II2;    ***£*  D,ia"  tor,  the  loop  tends  to  move  so  that  its  plane 

gram  of  the  Oscillograph. 

is  perpendicular  to  the   lines  of  magnetic 

flux  passing  from  N  to  S.  This  movement  of  the  loop  causes  the 
mirror  to  be  deflected,  the  deflection  is  proportional  to  the  current 
flowing  in  the  loop,  and  the  beam  of  light  reflected  from  the  mirror 
acts  as  a  pointer.  The  moving  parts  of  an  oscillograph  are  made 
very  light  so  that  the  deflection  of  the  mirror  is  always  in  time- 

*  The  inductance  of  the  armature  is  proportional  to  the  square  of  the  number  of 
conductors  per  slot. 


THE  ALTERNATING-CURRENT  GENERATOR 


127 


phase  with  the  current  flowing  in  the  current-carrying  loop,  and 
the  position  of  the  mirror  indicates  the  instantaneous  value  of  the 
current  flowing  in  the  loop. 

If  the  beam  of  light  reflected  from  the  mirror  M  is  directed  onto 
a  photographic  film  having  a  uniform  motion  at  right  angles  to 
the  movement  of  the  beam  of  light,  records  similar  to  those  in 
Fig.  in  are  obtained.  When  it  is  not  desired  to  make  photo- 
graphic records  a  second  mirror,  operated  by  a  small  synchronous 
motor,  is  rotated  or  vibrated  in  such  a  manner  that  the  reflection 
of  the  beam  of  light  on  a  stationary  screen  indicates  the  wave  form 
of  the  current  or  electromotive  force. 

6.  The  single-phase  alternating-current  generator.  —  In  the 
single-phase  alternating-current  generator,  the  armature  conduc- 
tors are  all  connected  in  series.  Fig.  113.  It  is  evident  from 
the  considerations  discussed  in 
Section  3,  that  the  value  of  k  is 
low  if  the  armature  conductors  are 
distributed  over  the  entire  arma- 
ture surface.  The  terminal  volt- 
age of  a  single-phase  alternator 
is  only  slightly  greater  when  the 
'  conductors  cover  the  entire  arma- 
ture surface  than  when  they  cover 
three-fourths  of  the  surface,  but 
the  weight  of  copper  and  the  arm- 
ature resistance  are  proportional 
to  the  surface  covered.  It  is, 
therefore,  common  practice  to  distribute  the  armature  conductors 
over  only  a  portion  of  the  surface  of  the  armature  core.  Because 
of  this  excess  of  materials,  a  single- phase  alternator  has  a  greater 
weight  and  is  less  efficient  than  a  polyphase  alternator  of  the  same 
voltage  and  output. 

Consider  an  armature  having  four  slots  per  pole  and  so  wound 
that  the  average  electromotive  force  induced  in  the  conductors 
in  each  slot  is  10  volts.     The  arc  subtended  by  this  winding  is  180 
degrees,  the  value  of  k  is  0.724,  and  the  effective  electromotive 
force  induced  in  this  section  of  the  armature  winding  is 
Ea  =  4  X  10  X  0.724 
=  28.96  volts. 


FIG.  113.    Elementary  Single-phase 
Armature  Winding. 


128 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


If  three  of  the  four  slots  in  the  above  armature  core  are  used,  the 
arc  subtended  is  135  degrees,  the  value  of  k  is  increased  to  0.893, 
and  the  effective  electromotive  force  induced  in  the  coils  is 

Ea  =  3  X  10  X  0.893 
=  26.79  volts. 

Thus  by  using  only  three  of  the  four  slots  per  pole,  the  electro- 
motive force  induced  in  the  armature  winding  is  reduced  only 
7.5  per  cent,  while  the  weight  of  copper  and  the  resistance  of  the 
armature  circuit  are  each  reduced  25  per  cent.  The  reactance  of 
the  armature  is  also  materially  reduced. 

7.  The  two-phase  alternating-current  generator.  —  In  the  two- 

phase  alternator  two  armature 
windings  are  wound  on  the  same 
core,  the  conductors  being  so 
placed  that  the  two  induced  elec- 
tromotive forces  are  in  quadrature. 
As  each  armature  winding  covers 
one-half  the  surface  of  the  core, 
the  arc  subtended  is  90  degrees, 
the  value  of  k  is  materially  higher 
than  for  the  single-phase  alterna- 
tor, the  material  is  used  more  ad- 
vantageously, and  a  larger  output 
is  obtained  from  a  given  weight. 

8.  The    three-phase    alternating-current    generator.  —  In   the 
three-phase  alternator  three  armature  windings  are  wound  on  the 
same  core  and  so  placed  that  the  induced  electromotive  forces  are 
1 20  degrees  out  of  phase.    Each  winding  covers  one-third  of  the 
surface  of  the  core,  and  the  arc  subtended  by  the  winding  is  60 
degrees. 

The  windings  of  three-phase  alternators  are  always  intercon- 
nected, the  star  connection  being  more  largely  used  than  the  delta.* 
By  connecting  a  single-phase  load  between  any  two  terminals,  a 
star-connected  three-phase  alternator  may  be  operated  as  a  single- 
phase  generator  in  which  the  active  armature  conductors  cover 
two-thirds  of  the  armature  surface. 

*  In  a  delta-connected  system  harmonics  cause  a  current  to  circulate  around  the 
delta  and  heat  the  conductors.  Also,  for  a  given  electromotive  force  a  delta-connected 
armature  requires  73  per  cent  more  conductors  than  does  a  star-connected  armature. 


FIG.  114.     Elementary  Two-phase 
Armature  Winding. 


THE  ALTERNATING-CURRENT  GENERATOR 


129 


9.  Armature  reaction.  —  As  in  the  continuous-current  dynamo, 
the  armature  current  of  an  alternator  affects  both  the  value  of  the 
flux  in  the  air  gap  and  its  distribution.  If  the  current  flowing 
in  the  armature  windings  is  in  phase  with  the  induced  electro- 


FIG.  lisa.  Elementary  Three-phase 
Armature  Winding  (Delta  Con- 
nected) . 


FIG.  iisb.  Elementary  Three-phase 
Armature  Winding  (Star  Con- 
nected). 


motive  force,  the  flux  is  distorted  only;  if  the  current  and  the 
electromotive  force  are  not  in  phase,  the  total  flux  is  increased  or 
decreased  as  the  current  leads  or  lags  behind  the  electromotive 
force. 

Referring  to  a  single-phase  two-pole  alternator  with  concentrated 
winding,  let 

N  =  the  number  of  series  conductors  on  the  armature, 
<t>  =  the  angle  by  which  the  current  leads  or  lags  behind  the 
electromotive  force. 


Then 


and 


e  =  Em  sin  co/, 


i  =  Im  sin  (ut  ±  </>). 


(7) 
(8) 


The  direction  of  the  flux  set  up  when  current  flows  in  the  arma- 
ture winding  is  at  right  angles  to  the  plane  of  the  coil,  and  its  value 
is  proportional  to  the  product  of  the  armature  current  and  the 
number  of  turns  in  the  winding.  Let  the  instantaneous  magneto- 


130  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

motive  force  in  ampere-turns  due  to  the  armature  winding  be 
represented  by  m.    Then 


t  \ 
m=--  (9) 

^_  JV7msin(co/dr<fr)_  ,    v 

2 

This  magnetomotive  force  may  be  resolved  into  components  at 
right  angles  to  each  other,  one  of  which  (m')  is  parallel  to  the  axis 
of  the  field  flux,  and  increases  or  decreases  the  flux  set  up  by  the 
field  windings;  the  other  component  (m"}  is  in  quadrature  with 
the  field  flux  and  changes  the  distribution  of  the  flux  in  the  air  gap 
(distorts). 

,       ]V/TOsin(co/:i=<£)  /     N 

m=-  ^coswf,  (n) 


.  /    N 

sin  co/.  (12) 

Expanding  equation  (n) 

(    } 


2 

NIm  (sin  co/  cos  co/  cos  <ft  j=  cos2  co/  sin  <ft) 
=  -  -  . 

2 

But 

av.  cos2  co/  =  0.5  (15) 

and 

av.  sin  co/  cos  co/  cos  <£  =  o.  (16) 

Therefore, 

f    ^ 
(17) 


i.c.,  the  average  magnetizing  magnetomotive  force  set  up  by  the 
armature  winding  of  a  single-phase  alternator  is  proportional  to 
the  product  of  the  maximum  value  of  the  current  flowing  in  the 
winding  and  the  sine  of  the  angle  of  phase  difference.  Similarly, 


av.  m    = 


/  QN 
(18) 


*  If  the  armature  winding  is  distributed,  the  effective  ampere-turns  tending  to  set 

Ni 
a  flux  at  right  angles  to  the  plane  of  the  coil  equals  — 

of  one-half  the  angle  over  which  the  winding  is  distributed. 


Ni 
up  a  flux  at  right  angles  to  the  plane  of  the  coil  equals  —    times  the  average  cosine 


THE  ALTERNATING-CURRENT  GENERATOR  131 

i.e.,  the  average  distorting  magnetomotive  force  set  up  by  the 
armature  winding  of  a  single-phase  alternator  is  proportional  to 
the  product  of  the  maximum  value  of  the  current  flowing  in  the 
winding  and  the  cosine  of  the  angle  of  phase  difference. 

The  direction  and  relative  magnitude  of  the  armature  magneto- 
motive force  of  a  single-phase  alternator  are  shown  in  Fig.  116  for 
different  positions  of  the  armature.  It  will  be  observed  that  the 


CO 


FIG.  1  1  6.    Armature  Magnetomotive  Force.    Single-phase  Alternator. 

direction  of  the  magnetomotive  force,  as  well  as  its  magnitude, 
varies  with  the  position  of  the  armature. 

In  polyphase  alternators  with  balanced  load,  the  armature  mag- 
netomotive force  is  constant  in  value  and  fixed  in  position  *  for  any 
given  armature  current  and  phase  angle.  For  a  two-phase  alter- 
nator, the  magnetomotive  force  set  up  by  one  winding  is 


mi  = 


and  that  set  up  by  the  other  winding  is 

_  A^m  sin  (a;/ dr  QO  ± 
2 


,    . 
(19) 


(20) 
(21) 


*  It  is  assumed  that  the  alternator  is  of  the  rotating  armature  type.  When  the 
armature  conductors  are  stationary,  the  flux  set  up  by  the  armature  winding  is  con- 
stant in  value  but  rotates  at  a  speed  proportional  to  the  frequency  of  the  currents 
flowing  in  the  armature  conductors.  This  rotation  of  flux  is  the  fundamental  principle 
of  the  induction  motor.  See  Chapter  13. 


132  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

Since  these  two  magnetomotive  forces  are  at  right  angles, 

M 


NIm  Vsin2  (<*t  =b  0)  +  cos2  (tat  d= 

2 


(a  constant), 


i.e.,  the  magnetomotive  force  set  up  by  the  armature  winding  of  a 
two-phase  alternator  is  constant,  and  equal  to  the  maximum  value 
of  the  magnetomotive  force  set  up  by  each  phase  winding. 

That  the  direction  of  the  magnetomotive  force  set  up  by  a  two- 
phase  armature  is  fixed  is  evident  from  the  following  consider- 
ations: At  the  instant  when  the  current  in  phase  A  is  maximum, 
the  armature  is  in  the  position  indicated  in  Fig.  nya,  the  current 


(0) 


(b) 


(e)  (f) 

FIG.  117.     Armature  Magnetomotive  Force.    Two-phase  Alternator. 

in  phase  B  is  zero,  and  the  magnetomotive  force  set  up  by  the  arma- 
ture is  at  right  angles  to  the  plane  of  coil  A.  It  is  represented, 
both  in  magnitude  and  in  direction,  by  M.  When  the  armature 
has  rotated  through  an  arc  of  30  degrees  (one-twelfth  of  a  revolu- 
ton),  the  coils  are  in  the  positions  indicated  in  Fig.  nyb,  and 


THE  ALTERNATING-CURRENT  GENERATOR  133 

the  magnetomotive  force  set  up  by  the  armature  is  the  geometric 
sum  of  the  magnetomotive  forces  set  up  by  coils  A  and  B. 
Similarly,  the  conditions  existing  when  the  armature  is  in  other 
positions  are  shown  for  a  half  revolution  of  the  armature  which 
brings  the  coils  into  their  first  position,  but  with  the  currents  flow- 
ing in  the  opposite  direction. 

That  the  magnetomotive  force  produced  by  the  armature  of  a 
three-phase  alternator  is  constant  in  value  and  fixed  in  space  may 
be  shown  in  a  similar  manner,  the  value  of  the  magnetomotive 
force  being  one  and  one-half  times  the  maximum  value  produced 
by  each  phase. 

10.  Voltage  characteristic.  —  From  Section  9  it  is  evident  that 
the  field  excitation  required  to  simultaneously  produce  rated  ter- 
minal voltage  and  rated  armature  current  varies  with  the  character 
of  the  load,  i.e.,  armature  reaction  decreases  or  increases  the  flux 


20          40  60          60  100          120         140 

PERCENT  OF  RATED  ARMATURE, CURRENT 

FIG.  118.    Voltage  Characteristics. 

set  up  by  a  given  field  current  as  the  armature  current  lags  behind 
or  leads  the  induced  electromotive  force.  Therefore,  the  rise  in 
voltage  as  the  armature  current  decreases,  is  greater  when  the  load 
circuit  is  inductive  than  when  it  is  non-inductive,  and  the  terminal 
voltage  may  decrease  as  the  armature  current  decreases,  if  the 
alternator  is  supplying  a  capacitive  load  circuit.  The  voltage 
characteristic  of  an  alternator  is  a  curve  showing  the  relations 
between  the  terminal  voltage  and  the  armature  current,  the 
speed  and  the  field  excitation  remaining  constant.  Curves  for 
inductive,  non-inductive  and  capacitive  load  circuits  are  shown  in 
Fig.  1 1 8. 


134 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


ii.  Regulation.  —  The  regulation  of  an  alternating-current 
generator  is  the  ratio  of  the  increase  in  voltage,  between  rated  load 
and  no  load,  and  the  voltage  at  full  load,  the  frequency  and  the 
field  excitation  remaining  constant. 

(no-load  voltage  —  full-load  voltage)  100 
full-load  voltage 


Per  cent  regulation 


(24) 


It  is  a  waste  of  considerable  energy,  and  often  inconvenient  or 
impossible,  to  determine  the  regulation  of  a  large  alternator  by 
an  actual  load  test.  The  regulation  of  an  alternator  is,  therefore, 
usually  calculated  from  no-load  tests.  The  data  necessary  for  the 
calculation  of  the  regulation  of  an  alternating-current  generator 

are:  (a)  the  resistance 
of  the  armature  circuit, 
(b)  the  open-circuit  sat- 
uration (magnetization) 
curve,  (c)  the  "  short- 
circuit"  curve,  (d)  the 
zero  power  factor  satu- 
ration curve. 

(a)  Armature  resist- 
ance. --To  determine 
the  resistance  of  the 
armature  winding,  con- 
nect the  armature  to 

FIG.  119.    Saturation,  Short-circuit  and  Synchro-    continuous  -  current 
nous  Reactance  Curves  for  the  Alternator.  mains    in    Series  with    a 

suitable  resistance,  so 
the  current  in  the  arm- 
ature circuit  may  be 
controlled.  The  resist- 
ance of  the  armature 

winding  is  the  ratio  of  the  voltage  between  its  terminals  and  the 

current  flowing  in  the  circuit. 


^ 

^ 

o 

^ 

-^ 

-^> 

2 

Ra 

ed 

Volt 

a^ 

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36V 

S 

^ 

1 

/ 

/ 

s> 

> 

14 

^2000 

t- 

/ 

/ 

/\ 

'L 

in 

j    - 
°,fior 

/ 

' 

/ 

j 

'/ 

K     ' 

>  '6°° 

^ 

/*s 

V 

} 

'/ 

b 

^ 

^10-^ 

/ 

g 

^ 

^ 

^ 

* 

§     o 

/ 

/ 

// 

S 

X. 

! 

/ 

1 

// 

V" 

^ 

-^ 

5  ' 

/ 

/ 

J 

V 

/ 

f 

s^ 

f 

100 

/^ 

"v 

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^ 

^ 

\ 

& 

•c 

I 

D 

Z 

0 

I 

0 

4 

0 

5 

0 

6 

0 

1 

) 

FIELD   AMPERES 


OA  open-circuit  saturation  curve. 

BC  approximate  zero  power  factor  curve. 

BC'  experimental  zero  power  factor  curve. 

BD  full-load  saturation  curve  at  unity  power  factor. 

OF  short-circuit  curve. 

EG  synchronous  reactance  curve. 


Ice 

(b)  The  open-circuit  saturation  curve.  —  The  saturation  curve 
shows  graphically  the  relations  between  the  electromotive  force  in- 
duced in  the  armature  winding  and  the  field  excitation.  The  data 


THE  ALTERNATING-CURRENT  GENERATOR        135 

from  which  the  saturation  curve  is  plotted  are  obtained  by  meas- 
uring the  terminal  voltage  of  the  armature  with  the  armature 
circuit  open,  for  different  field  currents,  the  frequency  being  main- 
tained at  the  rated  value.  Fig.  119. 

(c)  The  "short-circuit"  curve. — The  short-circuit  curve  shows 
graphically  the  relations  between  the  current  in  the  short-circuited 
armature  winding  and  the  field  excitation.  Taking  into  considera- 
tion the  rated  current  of  the  armature,  short  circuit  the  armature 
winding,  as  indicated  in  Fig.  120,  through  a  suitable  ammeter, 
and  determine  the  field  current  required  to  cause  different  current 


Field  Rheostat-          , ,  Field  Rheostat 

flfo 


To  Lucifer 


(a)  5 ingle  Phase  (b)  Three  Phase 

FIG.  1 20.     Connections  for  Short-circuit  Test. 

values  to  flow  in  the  armature  circuit,  the  rotating  parts  of  the 
alternator  being  driven  at  rated  speed.  Plot  the  curve  as  in  Fig. 
119.  The  impedance  of  the  armature  circuit  is  assumed  to  be 
constant  and  the  short-circuit  curve  is,  approximately,  a  straight 
line. 

(d)  The  zero  .power  factor  saturation  curve.  —  The  zero  power 
factor  saturation  curve  of  an  alternator  shows  the  relation  between 
the  terminal  voltage  and  the  field  excitation  when  the  power  factor 
of  the  load  circuit  is  zero,  and  rated  current  flows  in  the  armature 
circuit.  A  curve  approximating  the  zero  power  factor  curve  may 
be  constructed  from  data  obtained  when  the  load  on  an  alternator 
consists  of  idle-running  under-excited  synchronous  motors,  the 
power  factor  of  which  is  very  low. 

An  approximate  zero  power  factor  curve  may  also  be  con- 
structed from  the  open-circuit  saturation  curve  and  the  short- 
circuit  curve.  Draw  the  curve  BC,  in  Fig.  119,  parallel  to  the 
open-circuit  saturation  curve,  beginning  at  the  abscissa  which 
represents  the  field  excitation  required  to  cause  rated  current  to 
flow  in  the  short-circuited  armature.  In  alternators  having  high 
reactance,  high  saturation,  and  large  magnetic  leakage,  the  zero 
power  factor  curve  may  lie  before  BC,  as  indicated  by  the  line 
BC,  and  its  exact  location  must  be  determined  by  test. 


136  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

Regulation  by  the  electromotive-force  method.  —  In  calculating  the 
regulation  of  an  alternator  by  the  electromotive-force  method, 
it  is  assumed  that  the  electromotive  force  induced  in  the  armature 
is  the  vector  sum  of  two  quadrature  electromotive  forces,  one  equal 
to  the  product  of  the  armature  current  and  the  total  resistance  of 
the  circuit,  including  that  of  the  armature;  the  other  equal  to  the 
product  of  the  armature  current  and  the  total  reactance  of  the 
circuit. 


Ea  =  ia  VR*  +  x2  (26) 


cos  0  +  Ralrf  +  (E  sin  <£  +  XJa)\  (27) 

when  Ea  =  the  electromotive  force  induced  in  the  armature, 
E  =  the  terminal  electromotive  force  of  the  generator, 
Ra  =  the  resistance  of  the  armature  circuit, 
Xa  =  the  reactance  of  the  armature  circuit  (the  synchronous 

reactance*  of  the  alternator  armature), 
I  a  =  the  current  flowing  in  the  armature  circuit, 
cos  $  =  the  power  factor  of  the  load  circuit. 

The  synchronous  reactance  of  the  armature  circuit,  which  can- 
not be  measured  directly,  must  be  calculated.  The  voltage  drop 
in  the  armature  is  the  vector  sum  of  the  resistance  voltage  and 
the  reactance  voltage. 

ES  =  V(Rjay  +  (xjaY-  (28) 

Dividing  equation  (28)  by  Ia, 


za=  VRJ  +  XJ,  (29) 


and  Xa  =  Vza2  -  Ra2.  (30) 

A  consideration  of  the  open  circuit  and  the  zero  power  factor 
saturation  curves,  shows  that  the  synchronous  reactance  of  an  al- 

*  The  synchronous  reactance  of  an  alternator  armature  is  the  combined  effect  of :  (a) 
the  inductance  of  the  armature  windings,  (6)  the  flux  set  up  by  the  current-carrying 
armature  conductors. 

As  shown  in  Section  9,  the  armature  currents  of  a  polyphase  alternator  set  up  a  flux 
which  is  fixed  in  position,  and  across  which  the  armature  conductors  move.  The 
electromotive  force  induced  in  the  conductors  by  the  armature  flux  reduces  the 
terminal  voltage  of  the  alternator  and  makes  the  regulation  poorer  in  much  the  same 
manner  as  does  the  inductance  of  the  windings. 

A  similar  effect  takes  place  in  the  single-phase  alternator  because  of  the  varying 
value  of  the  flux  threading  the  armature  coil. 


THE  ALTERNATING-CURRENT  GENERATOR        137 

ternator  armature  is  not  constant,  but  decreases  as  the  excitation 
increases.  It  is  also  affected  by  the  phase  relation  of  the  current 
and  the  electromotive  force  induced  in  the  armature  winding,  i.e., 
by  the  power  factor  of  the  load  circuit. 

Substituting  in  equation  (27)  the  full-  (rated)  load  armature  cur- 
rent, the  no-load  voltage  is  calculated  and 

,  , .          (Ea  —  E)  100  f    x 

Per  cent  regulation  =  -     — — —    -  •  (31) 

E 

Example.  —  A  single-phase  alternator  has  the  following: 

E  =   2300,  Ia  =   100,  Ra  =   I,  Xa  =   10. 

Find  the  per  cent  regulation  when  the  power  factor  of  the  load 
circuit  is  unity. 

Solution.  

£«  =  V(23oo  +  ioo)2  +  (iooo)2 
=  2600  volts. 

-p.       ,  ,.  2600  —  2300 

Regulation  =  - 

2300 

=  13  per  cent. 

Regulation  by  the  magnetomotive-force  method.  —  In  calculating 
the  regulation  of  an  alternator  by  the  magnetomotive  force  method 
it  is  assumed  that  the  voltage  induced  in  the  armature  windings 
is  due  to  quadrature  fields,  one  equal  to  that  required  to  produce 
the  total  non-inductive  drop,  the  other  equal  to  that  required  to 
produce  the  total  wattless  component  of  electromotive  force. 

From  the  saturation  curve  find  the  field  current  //  required  to 
induce  in  the  armature  winding  an  electromotive  force  equal  to 
the  total  non-reactive  drop  (E  cos  <£  +  RJa)  of  the  circuit  at 
rated  load;  from  the  same  curve  find  the  field  current  //'  required 
to  induce  in  the  armature  winding  an  electromotive  force  equal  to 
the  total  reactive  drop  (E  sin  <j>  +  XJa)  of  the  circuit  at  rated  load. 

The  field  current  //  required  to  produce,  simultaneously,  the  ter- 
minal voltage  E  and  the  current  Ia  in  a  circuit  the  power  factor  of 
which  is  cos  <£,  is 

It  =  V(//)» +  (/,")'.  (32) 

,    From  the  saturation  curve  find  the  electromotive  force  Ea  in- 


138  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

duced  in  the  armature  winding  when  current  //  flows  in  the  field 
windings. 

Per  cent  regulation  =  ^--?—       —  .  (33) 

h, 

Example.  —  Find,  by  the  magnetomotive  force  method,  the  reg- 
ulation of  the  2  300-  volt,  single-phase  alternator,  specified  above. 
Solution.  —  From  Fig.  119, 

//  =  44.5  amperes, 

and 

//'  =12.5  amperes. 

Therefore,  _ 

If    =  V(44.5)2  +  (i2.5)2 
=  46.22  amperes. 

The  induced  electromotive  force  is,  from  the  open-circuit  satura- 
tion curve,  2430  volts,  and  the  regulation  is 


2300 

=  5.2  per  cent. 

Regulation  by  the  A.I.E.E.  method.  —  The  electromotive  force 
induced  in  the  armature  winding  of  an  alternator  is  the  vector 
sum  of  the  terminal  voltage  and  the  voltage  drop  in  the  armature 
circuit.  Therefore,  the  drop  in  the  armature,  at  zero  power  factor 
and  rated  armature  current,  is  the  difference  between  the  ordinates 
of  the  open-circuit  and  the  zero  power  factor  saturation  curves. 

For  field  excitation,  Oc,  the  induced  electromotive  force  is  ca, 
and  the  drop  in  the  armature  is  ba.  Fig.  119.  For  any  other 

power  factor,  and  the  same  field  excita- 
tion, the  terminal  voltage  is  the  vector 
difference  of  the  induced  electromotive 
force  and  the  internal  drop.  Fig.  121. 
By  calculating  the  terminal  electro- 
motive forces  for  different  field  exci- 
tations but  constant  power  factor,  the 
load  saturation  curve  BD,  in  Fig.  119, 
may  be  plotted.  From  this  curve  and 
the  open-circuit  saturation  curve,  the 
electromotive  force  induced  in  the  armature  when  the  alternator 
is  operating  at  rated  voltage  and  armature  current,  and  with 


THE  ALTERNATING-CURRENT  GENERATOR        139 

the  specified  power  factor,  is  determined,  and  the  regulation  cal- 
culated. 

From  Fig.  119,  the  induced  voltage  is  2490  when  rated  current 
flows  in  the  armature  circuit,  the  terminal  voltage  is  2300,  and 
the  power  factor  of  the  load  circuit  is  unity.  The  regulation  is, 
therefore, 

=  !&-  X  100 

/  2300 

=  8.3  per  cent. 

12.  Limits  of  regulation.  —  From  the  above  numerical  examples 
it  is  evident  that  the  electromotive  force  method  and  the  magneto- 
motive-force method  do  not  give  the  same  result,  these  calculations 
serving  only  to  establish  the  limits  within  which  the  actual  regu- 
lation lies.     If  the  saturation  curve  was  a  straight  line,  the  actual 
regulation  would  be  determined  by  either  of  these  methods.     Since 
the   no-load  voltage,   as   calculated    by   the    electromotive-force 
method,  is  larger  than  that  obtained  by  an  actual  load  test,  this 
method  is  termed  "  pessimistic. "    The  no-load  voltage,  as  obtained 
by  the  magnetomotive  force  method,  is  smaller  than  that  obtained 
by  an  actual  load  test,  and  this  method  is  termed  "optimistic. "     It 
is  worthy  of  note  that,  in  a  well-designed  alternator,  the  optimistic 
method  gives  a  closer  approximation  to  the  actual  regulation  than 
does  the  pessimistic  method. 

The  A.I.E.E.  method  for  the  calculation  of  alternator  regula- 
tion is  essentially  empirical,  but  gives  results  which  approximate 
very  closely  those  obtained  by  test,  and  its  use  is  recommended  in 
preference  to  either  the  electromotive-force  or  the  magnetomotive- 
force  methods. 

13.  Graphical  determination  of  term- 
inal voltage.  —  In  Fig.   122,  using  0 
as  a  center  and  with  a  radius  propor- 
tional to  the  rated  electromotive  force 
of  the  alternator,  strike  an  arc  ECD. 
Through  0  draw  the  current  vector  Ia. 

Lay  off  OA  proportional  to  the  full-  FlG-  I22-  Alternator  Regulation, 
load  resistance  drop  RJa  of  the  armature  winding,  and  AB  propor- 
tional to  the  full-load  reactance  drop  XJa  of  the  armature  cir- 
cuit. Draw  OC  to  its  intersection  with  the  arc  ECD,  making  the 
angle  COI  such  that  its  cosine  is  equal  to  the  power  factor  of  the 


140 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


load  circuit.  BC  is  proportional  to  the  generated  or  no-load 
voltage. 

With  B  as  a  center  and  a  radius  equal  to  BC,  strike  a  second  arc 
FCG.  Divide  OB  into  any  number  of  equal  parts  and  lay  off  equal 
spaces  beyond  O,  the  spaces  representing  percentages  of  the  rated 
capacity  (armature  current)  of  the  alternator.  From  these  points 
draw  lines,  parallel  to  OC,  to  their  intersection  with  the  arc  FCG. 
The  lengths  of  these  parallel  lines  are  proportional  to  the  terminal 
voltages  at  the  given  percentages  of  rated  load,  and  from  them  the 
voltage  characteristic  may  be  constructed,  or  the  regulation  cal- 
culated. 

14.  The  Tirrill  Regulator.  —  The  operation  of  the  Tirrill  regu- 
lator, when  applied  to  an  alternator,  is  essentially  the  same  as  for  the 


Main  Contacts 


A.C.  field  Rheostat  A-c-  Generator 

FIG.  123.    Tirrill  Regulator  for  Alternators. 

continuous-current  generator.*  An  elementary  diagram  showing 
the  application  of  a  Tirrill  regulator  to  an  alternator  is  shown  in 
Fig.  123.  The  field  rheostat  of  the  exciter  is  periodically  short- 
circuited  by  the  closing  of  the  main  contacts,  the  length  of  time 
during  which  the  short  circuit  exists  depending  on  the  position  of 
the  main  contact.  The  position  of  this  contact  is  controlled  by  the 
alternating-current  magnet,  i.e.,  by  the  voltage  of  the  alternating- 
current  system.  The  function  of  the  compensating  winding  is  to 
increase  the  terminal  voltage  as  the  load  increases,  the  effect  of  the 
compensating  winding  being  proportional  to  the  line  current. 

Assuming  the  main  contacts  to  be  open,  the  relay  contacts 
are  held  open  by  the  differentially  wound  relay  magnet,  and  the 
voltage  of  both  the  exciter  and  the  alternator  falls  off.     The  re- 
duced voltages  allow  the  main  contacts  to  close.     Closing  the 
*  See  Chapter  4,  Section  10. 


THE  ALTERNATING-CURRENT  GENERATOR        141 

main  contacts  demagnetizes  the  relay,  closes  the  relay  contacts, 
and  short  circuits  the  exciter-field  rheostat.  The  increased  field 
excitation  increases  the  electromotive  force  between  the  terminals 
of  the  control  magnets,  and  causes  the  main  contacts  to  open. 

15.  The  losses  in  an  alternating-current  generator.  —  The  losses 
in  an  alternator  are:  (a)  armature  copper  losses,  (b)  field  copper  losses, 

(c)  windage  and  friction,  (d)  iron  (core)  losses,  (e)  stray  load  losses. 

(a)  Armature  copper  losses.  —  The  copper  loss  in  the  armature 
winding  is  that  due  to  the  resistance  of  the  armature  conductors, 
and  increases  as  the  square  of  the  armature  current. 

(b)  Field  copper  losses.  —  The  field  copper  losses  are  those  due 
to  the  resistance  of  the  field  winding  and  are  proportional  to  the 
square  of  the  field  current.     If  the  terminal  voltage  is  to  remain 
constant,  the  field  excitation  of  an  alternator  must  be  increased  as 
the  load  increases  to  compensate  for  armature  reaction  and  resist- 
ance drop,  and  the  field  losses  increase  with  the  load. 

(c)  Windage  and  friction.  —  The  windage  and  friction  losses  are 
those  due  to:   (i)  friction  between  the  shaft  and  the  bushings, 
(2)  friction  between  the  brushes  and  the  rings,  (3)  the  resistance 
offered  by  the  air  to  the  movement  of  the  rotating  parts.     These 
losses  are  constant  for  any  given  speed. 

(d)  Iron  losses.  —  The  iron  losses  are  due  to:  (i)  hysteresis  in  the 
iron  of  the  armature  core,  (2)  eddy  currents  in  the  armature  core. 
The  iron  losses  do  not  vary  greatly  and  are  usually  considered 
constant. 

(e)  Stray  load  losses.  —  The  stray  load  losses,  so-called  for  lack  of 
.a  better  name,  include  all  the  losses  not  included  in  (a),  (£),  (c)  and 

(d)  and  such  changes  in  (c)  and  (d)  as  take  place  when  the  alter- 
nator is  loaded.     The  stray  load  losses  are  primarily  additional  iron 
losses  which  are  due  to  distortion  of  the  magnetic  field  as  the  arma- 
ture current  increases;  to  eddy-current  losses  in  the  armature  con- 
ductors, etc.;  and  are  directly  proportional  to  the  armature  current. 

Determination  of  the  losses  in  an  alternating-current  generator*  — 
The  losses  of  an  alternator  are  determined  as  follows: 

Armature  copper  losses.  —  The  copper  loss  in  the  armature  con- 
ductors is  equal  to  the  product  of  the  resistance  of  the  armature 
circuit  and  the  square  of  the  armature  current. 

Pa   =  RaV.  fo) 

*  See  the  Standardization  Rules  of  the  American  Institute  of  Electrical  Engineers. 


142  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

Field  copper  loss. —   . 

P/ =  */[(//)'  +  (/"/)*!,  (35) 

when  Rf  =  the  resistance  of  the  field  circuit, 

//  =  the  field  current  (taken  from  the  saturation  curve)  re- 
quired to  produce  rated  voltage  at  no  load. 
I/'  =  the  field  current  (taken  from  the  short-circuit  curve) 
required  to  produce  a  given  current  in  the  short- 
circuited  armature. 

Windage  and  friction.  —  Drive  the  alternator  at  its  rated  speed 
but  without  field  excitation,  and  determine  the  input  to  the  driving 
motor.  Determine  the  losses  in  the  motor  and  subtract  them  from 
the  total  input.  The  difference  is  the  loss  due  to  windage  and 
friction  of  the  alternator. 

Pw&f  =  Motor  input  —  motor  losses.  (36) 

Iron  losses.  —  Drive  the  alternator  at  its  rated  speed  and  with 
such  field  excitation  as  gives  rated  voltage  at  the  armature  terminals. 
The  total  input  to  the  driving  motor  is  the  sum  of  the  losses  in  the 
motor,  the  windage  and  friction  of  the  alternator,  and  the  iron  losses 
of  the  alternator. 

Pi  =  Motor  input — motor  losses  —  Ww  &/.  (3  7) 

Stray  load  losses.  —  Drive  the  alternator  at  its  rated  speed  with 
the  armature  winding  short-circuited,  and  with  such  field  excitation 
as  produces  the  desired  current  in  the  armature  circuit.  The  in- 
put to  the  driving  motor  is  equal  to  the  sum  of  the  motor  losses, 
the  windage  and  friction  of  the  alternator,  the  copper  losses  in  the 
alternator  armature  winding,  and  the  load  losses. 

Pi  =  Motor  input  —  motor  losses  —  Ral<?  —  Ww&/.      (38) 

The  stray  load  losses  as  measured  under  short  circuit  are  greater 
than  when  the  alternator  is  operating  under  load  because  of  the 
low-power  factor,. practically  zero,  of  the  short-circuited  armature. 
Experiment  shows  that  when  one-third  the  stray  load  loss  as  de- 
termined by  the  above  method,  is  used  in  efficiency  calculations, 
the  results  approximate  actual  load  tests.  This  value  is,  there- 
fore, recommended  by  the  Standardization  Rules  of  the  American 
Institute  of  Electrical  Engineers. 


THE  ALTERNATING-CURRENT  GENERATOR 


143 


Fig.  124  shows  graphically  the  losses  in  an  alternator,  which  are 
so  plotted  that  the  total  loss  for  any  given  armature  current  is  ob- 
tained by  reading  the  ordinate  of  the  top  curve  corresponding  to 
the  given  armature  current. 


12 


25  50          75          100          IZ5          150 

PER  CENT  OF  RATED  ARMATURE  CURRENT 

FIG.  124.    Alternator  Losses. 


ZOO 


1 6.  Efficiency.  —  The  efficiency  of  an  alternator  is  calculated 
from  the  following  equation: 

(output)  IPO 


Per  cent  efficiency 


output  +  losses 


(39) 


17.  Parallel  operation  of  alternators.  —  Parallel  operation  of 
alternators  is  desirable  because  of  the  greater  efficiency  obtainable 
and  the  assurance  of  continuity  of  service,  as  explained  for  contin- 
uous-current generators. 

Alternating-current  generators  which  are  to  operate  in  parallel 
should:  (a)  have  equal  terminal  voltages,  (b)  have  the  same  fre- 
quency, (c)  attain  their  maximum  positive  values  of  electromotive 
force  at  the  same  instant,  (d)  have  similar  electromotive-force 
waves. 

If  any  one  of  the  above  conditions  does  not  exist,  a  current  flows 
around  the  local  circuit  formed  by  the  armatures  of  the  alternators, 
increases  their  heating,  reduces  the  capacity  of  the  alternators,  and 
often  causes  other  disturbances  in  the  system. 


144  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

If  the  voltages  are  equal,  the  frequencies  the  same,  the  electro- 
motive-force waves  similar,  and  the  maximum  positive  electromotive 
forces  attained  at  the  same  instant,  the  electromotive  forces  are 
at  all  times  equal  and  opposed,  and  current  does  not  flow  around 
the  local  circuit  formed  by  the  armatures  of  the  generators. 

If  the  voltages  are  unequal,  a  current,  the  magnitude  of  which  is 
proportional  to  the  difference  of  the  voltages,  flows  between  the 
alternators. 

.'•.  •        ••-SAfs.  (40) 


when  Z  =  the  impedance  of  the  local  circuit.  This  equalizing 
current  leads  the  electromotive  force  of  one  alternator  and  lags 
behind  that  of  the  other.  Fig.  125.  Except  for 
£a  the  resistance  loss,  this  current  is  wattless  and  its 
£«  effect  is  to  increase  the  magnetization  of  one  alter- 

FIG.  125.  nator  and  decrease  that  of  the  other  (Section  9), 
so  that  the  indicated  voltage  of  the  system  is  the  mean  of  the 
voltages  of  the  generators  when  operated  separately. 

If  the  frequencies  are  not  the  same,  the  instantaneous  voltages 
are  not  equal  at  all  times,  and  equalizing  currents  flow  between 
the  alternators. 

If  the  electromotive-force  waves  are  not  similar,  or  do  not  attain 
their  maximum  at  the  same  instant,  the  instantaneous  values  are 
unequal,  and  equalizing  currents  flow  between  the  generators  as 
described  above.  If  the  wave  forms  differ  greatly  or  the  alterna- 
tors are  out  of  phase  opposition  by  more  than  a  few  degrees,  the 
equalizing  currents  are  usually  so  large  as  to  operate  the  protec- 
tive devices  (fuses  or  circuit  breakers)  which  should  be  placed  in 
every  circuit. 

Therefore,  alternators  whose  operating  characteristics  or  wave 
forms  differ  radically  should  not  be  connected  in  parallel,  nor  is  it 
desirable  to  connect  in  parallel  alternators  whose  prime  movers 
have  materially  different  speed  characteristics. 

Synchronizing  and  division  of  load.  —  The  process  of  regulating 
the  voltage,  the  frequency  and  the  phase  relation  of  an  alternator 
so  that  it  may  be  connected  in  parallel  with  other  alternators,  is 
termed  "  synchronizing.  " 

The  connections  between   two   three-phase  alternators  and  a 


THE  ALTERNATING-CURRENT  GENERATOR 


145 


common  load  circuit  are  shown  in  Fig.  126.  If  A  is  carrying  the 
load  and  it  is  desired  to  connect  B  to  the  system,  start  B  and  regu- 
late its  field  excitation  and  the  speed  of  its  prime  mover  until: 

(a)  The  indication  of  its  voltmeter  is  equal  to  or  slightly  greater 
than  the  voltage  across  the  bus  bars. 


f    -*V Prime  Mover 


Generators  «5  -Generator  A 

FIG.  126.    Parallel  Operation  of  Alternators. 

(6)  The  lamps  connected  across  the  open  switch  S  are  dark  for 
several  seconds  at  a  time. 

Close  the  switch  S  quickly  at  a  time  when  the  lamps  are  dark. 

The  alternators  are  now  in  parallel  but  A  is  still  supplying  prac- 
tically all  the  load  current.  The  division  of  the  load  between  two 
alternators  operating  in  parallel  cannot  be  adequately  explained 
without  reference  to  the  principles  of  the  synchronous  motor,  so 
that  only  a  statement  of  facts  will  be  attempted  here. 

(af)  Changing  the  field  excitation  causes  a  wattless  current  to 
flow  between  the  armatures  without  changing,  to  any  appreciable 
extent,  the  distribution  of  the  load. 

(&')  The  load  distribution  on  alternators  operating  in  parallel  is 
changed  only  by  changing  the  relative  torques  of  the  prime  movers. 

The  lamps  in  Fig.  126  serve  the  double  purpose  of  indicating 
when  the  correct  phase  relations  are  obtained,  and  of  limiting  the 
flow  of  current  between  the  armatures  during  the  synchronizing 
period.  With  the  connections  indicated  in  Fig.  126,  all  lamps  are 
either  dark  or  bright  at  the  same  instant.  It  is  possible  to  make 
other  lamp  connections.  When  the  lamps  are  arranged  in  a  circle 
and  two  sets  are  cross-conneted  between  lines  A  and  B,  the  light 
appears  to  travel  around  the  circle,  its  direction  of  apparent  rota- 
tion depending  on  whether  the  speed  of  the  incoming  machine  is 
too  fast  or  too  slow. 

Because  synchronizing  lamps  give  only  an  approximate  indication 
of  the  time  when  the  switch  should  be  closed,  the  lamps  being  dark 


146 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


even  when  a  considerable  difference  of  potential  exists  between 
their  terminals,  mechanical  devices  known  as  " synchroscopes"  or 
"synchronism  indicators"  have  been  devised  and  are  in  extensive 


use.' 


18.  Ratings.  —  Alternators  are  rated  in  kilovolt-amperes  (kv-a.) 
rather  than  in  kilowatts,  because  the  power  output  of  a  given  alter- 
nator is  dependent  on  the  power  factor  of  the  load  circuit;  the  maxi- 


NS. 


s/Li* 


KM 


0.9 


0.8 


04 


0.3 


02 


07          0.6          05 

POWER   FACTOR 

FIG.  127.  Curves  Showing  the  Relation  between  Kilowatts,  Kilovolt-amperes  and 
Power  Factor,  the  Voltage  and  the  Armature  Current  of  the  Alternator  Remain- 
ing Constant. 

mum  load,  for  the  allowable  temperature  rise,  is  a  function  of  the 
armature  current.  The  kilowatt  output  of  a  generator  may  vary 
over  wide  limits,  by  a  variation  of  the  power  factor  of  the  load 
circuit,  the  current  output  remaining  constant.  Fig.  127. 

The  output  of  an  alternator  is  limited  by  the  allowable  heating 
of  the  armature,  and  by  saturation  in  the  iron  parts  of  the  mag- 
netic circuit.  Alternators  have  no  commutators,  and  are  not 
troubled  with  sparking. 

CHAPTER  VIII  —  PROBLEMS 

i.  The  armature  winding  of  a  single-phase  alternator  consists  of  1764  con- 
ductors (series)  which  are  distributed  uniformly  over  the  entire  armature  surface, 
and  in  each  of  which  an  average  electromotive  force  of  2  volts  is  induced.  There 
are  six  slots  per  pole.  Find:  (a)  the  terminal  voltage  of  the  alternator  at  zero 
load,  (b)  the  maximum  value  of  armature  magnetization  (in  effective  ampere- 

*  See  Chapter  17,  Section  9. 


THE  ALTERNATING-CURRENT  GENERATOR 


147 


turns)  when  the  current  output  of  the  generator  is  100  amperes,  (c)  the  mag- 
netizing and  the  distorting  components  of  armature  reaction  when:  (i)  the 
current  and  the  electromotive  force  are  in  phase,  (2)  the  current  lags  10  degrees 
behind  the  electromotive  force,  (3)  the  current  leads  the  electromotive  force 
by  15  degrees. 

2.  Same  as  Problem  i  except  the  armature  conductors  are  distributed  over 
f  of  the  armature  surface. 

3.  Same  as  Problem  i  except  the  armature  conductors  are  star-connected, 
and  the  current  in  each  line  is  100  amperes  (3-phase). 

4.  Same  as  Problem  3  except  the  armature  conductors  are  delta-connected 
(3-phase). 

5.  A  test  on  a  23oo-volt,  23o-kv.  a.,  single-phase  alternator  gave  the  follow- 
ing: 

ON  OPEN  CIRCUIT 


Field  current 

Terminal  voltage 

10 

500 

20 

1000 

30 

1500 

40 

1950 

50 

2250 

60 

2500 

70 

2650 

ON  SHORT  CIRCUIT 


Field  current 

Armature  current 

6.2S 

25 

12.50 

50 

18.75 

75 

25.00 

100 

Resistance  of  armature  =  2  ohms. 

Plot  the  saturation  curve  and  the  short-circuit  curve,  and  calculate,  by  the 
electromotive  force  method,  the  percentage  regulation  of  the  alternator  at :  (a) 
unity  power  factor,  (6)  85  per  cent  power  factor  lagging,  (c)  85  per  cent  power 
factor  leading. 

6.  Same  as  Problem  5  except  regulation  percentages  are  to  be  calculated  by 
the  magnetomotive  force  method. 

7.  Same  as  Problem  5  except  regulation  percentages  are  to  be  calculated 
by  the  A.I.E.E.  method. 

8.  Using  the  graphical  method  outlined  in  Section  1 2,  determine  the  terminal 
voltages  of  the  alternator  in  Problem  5  when  the  power  factor  of  the  load  circuit 
is  unity,  and  the  following  currents  flow  in  the  armature  winding:  o,  25,  50,  75, 
125,  150  amperes.    Plot  the  voltage  characteristic. 

9.  Same  as  Problem  8  except  the  power  factor  of  the  load  circuit  is  85  per 
cent  lagging. 


148 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


10.  Same  as  Problem  8  except  the  power  factor  of  the  load  circuit  is  85  per 
cent  leading. 

11.  A  test  of  a  66oo-volt,  looo-kv.-a.  3 -phase,  star-connected  alternator  gave 
the  following: 

ON  OPEN  CIRCUIT 


Field  ampere- 
turns 

Phase,  voltage 

2,000 

1071 

4,000 

2175 

6,000 

2990 

8,000 

3392 

10,000 

3534 

12,000 

3841 

14,000 

3979 

16,000 

4082 

18,000 

4163 

20,000 

4225 

22,000 

4280 

24,000 

4330 

An  excitation  of  4310  ampere-turns  is  required  when  full-load  (rated)  current 
flows  in  the  short-circuited  armature  windings. 

Armature  resistance  (measured  line  to  neutral)  =  i  ohm. 

Plot  the  magnetization  curve. 

Calculate  the  percentage  regulation  at  unity  power  factor  by:  (a)  the  electro- 
motive force  method,  (b)  the  magnetomotive  force  method,  (c)  A.I.E.E.  method. 

12.  Same  as  Problem  11  except  the  power  factor  of  the  load  circuit  is  90  per 
cent  lagging. 

13.  Same  as  Problem  11  except  the  power  factor  of  the  load  circuit  is  90  per 
cent  leading. 


CHAPTER  IX 
THE  SYNCHRONOUS  MOTOR 

i.  Structure.  —  Structurally,  the  synchronous  motor  is  identical 
with  the  alternating-current  generator,  and  the  same  machine  may 
be  used  as  a  generator  or  as  a  motor. 


(a)  Stator  (Armature). 


(b)  Rotor  (Field). 
FIG.  128.    The  Synchronous  Motor.    (Westinghouse.) 

2.  Principles  of  operation.  —  From  the  fundamental  principles 
explained  in  Chapter  2,  it  is  evident  that  a  conductor  carrying  an 
alternating  current  may  be  made  to  move  continuously  in  one 

149 


150  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

direction  if  the  direction  of  the  flux  through  which  it  moves  is  re- 
versed at  the  same  instant  that  the  direction  of  the  current  in  the 
conductor  is  reversed. 

Consider  the  looped  conductor  in  Fig.  129.  When  the  current 
is  flowing  away  from  the  observer  in  the  part  under  the  north  pole, 

and  towards  the  observer  in  the  part 
under  the  south  pole,  the  direction  of 
motion  is,  from  Fleming's  left-hand  rule, 
as  indicated  by  the  arrow.  This  direc- 
tion of  motion  is  made  continuous  by 
reversing  the  direction  of  the  current  in 
the  loop  when  the  plane  of  the  loop  is  in 
FlG*  I29>  the  vertical  position,  i.e.,  the  number  of 

revolutions  per  second  made  by  the  loop  must  be  equal  to  the  fre- 
quency of  the  alternating  current,  and  the  change  in  the  direction  of 
the  current  must  be  made  when  the  conductors  are  approximately 
midway  of  the  arc  between  the  poles.  Under  these  conditions,  the 
relations  between  the  field  flux  and  the  current  in  the  moving  con- 
ductor are  fixed,  and  the  torque  is  exerted  in  one  direction  only. 
Therefore,  if  the  frequency  of  the  supply  circuit  is  constant,  the 
speed  of  a  synchronous  motor  is  constant,  but  the  motor  is  not 
self-starting. 

3.  Torque-load  adjustment.  —  It  is  an  easily  demonstrated 
experimental  fact  that  when -the  driving  torque  of  an  alternating- 
current  generator,  opera  ting,  in  parallel  with  other  generators,  is 
reduced  to  zero,  its  moving  parts  continue  to  rotate  at  the  same 
speed,  and  that  the  direction  of  the  current  in  the  armature  windings 
is  reversed.  When  the  load  on  a  continuous-current  shunt  motor 
increases,  the  speed  of  the  armature  decreases  until  the  counter- 
electromotive  force  is  reduced. to  such  a  value  that  the  current 
necessary  to  produce  the  required  torque  flows  in  the  armature 
circuit.  Since  the  speed  of  a,  synchronous  motor  is  fixed  by  the 
frequency  of  the  system  to  which  it  is  connected,  the  increased 
armature  current  (torque)  must  be  produced  by  other  means. 

A  synchronous  motor,  like  a  continuous-current  motor,  generates 
a  counter-electromotive  force  which  is  proportional  to  the  speed, 
and  is  dependent  on  the  field  excitation.  The  current  in  the  arma- 
ture of  a  synchronous  motor  may,  then,  be  changed  by  changing 
the  field  excitation,  but  it  would  be  practically  impossible  to  oper- 


THE   SYNCHRONOUS  MOTOR  151 

ate,  commercially,  a  motor  the  field  current  of  which  must  vary 
with  the  load.  Consider  the  conditions  when  the  counter-electro- 
motive force  of  the  motor  is  equal  to  the  applied  electromotive 
force.  As  long  as  the  phase  difference  of  these  equal  electro- 
motive forces  is  180  degrees,  no  current  flows  in  the  armature  of 
the  motor,  and  no  torque  is  developed  in  it.  Consequently,  the 
rotating  parts  tend  to 
stop,  and  the  angle  of 
phase  difference  becomes 
greater  than  180  degrees. 
Let  the  phase  rela- 
tions of  the  two  equal 
electromotive  forces  be 
as  indicated  in  Fig.  130. 
A  current  proportional 
to  the  resultant  elec7 
tromotive  force  Er  flows  FlG-  T30- 
in  the  armature  of  the 


Mechanical  and  .Vector  Relations  of  an 
Alternator  and  a  Synchronous  Motor. 


motor,  and  the  angle  between  the  resultant  electromotive  force  and 
the  current  is  j3,  the  tangent  of  which  is  equal  to  --v5  • 


=  tan-1     - 


when  .Y0  =  the  synchronous  reactance  of  the  armature  circuit, 
Ra  =  the  resistance  of  the  armature  circuit. 

Since  Ra  is  always  small  as  compared  to  Xa,  the  angle  ft  is  usually 
greater  than  80  degrees  but  can  never  equal  90  degrees. 
The  input  to  the  motor  under  the  above  conditions  is 

~D  J7  T  (    \ 

If  the  load  increases,  the  angle  between  the  applied  and  the  counter- 
electromotive  forces  will,  evidently,  increase  until  equilibrium  is 
reestablished,  i.e.,  with  constant  field  excitation,  the  torque  of  a 
synchronous  motor  is  dependent  on  the  phase  relations  of  the  ap- 
plied and  the  counter-electromotive  forces,  and  the  angle  between 
them  is  automatically  adjusted  as  the  load  changes. 

4.   Starting.  —  A  synchronous  motor  is  not  self-starting  because 
the  periodical  reversal  of  the  direction  of  the  current  in  the  station- 


152  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

ary  armature  conductors  produces  a  torque  which  periodically 
reverses,  and  tends  to  produce  rotation  of  the  field  structure  first 
in  one  direction,  then  in  the  other.  The  commercial  methods  of 
starting  a  synchronous  motor  are:  (a)  by  means  of  an  auxiliary 
motor,  (6)  as  an  induction  motor. 

(a)  By  an  auxiliary  motor.  —  The  starting  motor  is  a   small 
motor,  either  shunt  or  induction,  direct-connected   or  belted  to 
the   shaft   of   the   synchronous   machine.     When   a   synchronous 
motor  is  started  in  this  way,  it  is  a  generator  and  must  be  syn- 
chronized *  before  it  is  connected  to  the  supply  circuit.     After  the 
synchronous  motor  armature  is  connected  to  the  alternating-cur- 
rent supply  circuit,  the  circuit  of  the  starting  motor  is  opened.     In 
case  the  synchronous  machine  is  provided  with  a  separate  exciter 
and  a  source  of  continuous  current  is  available,  the  exciter  may 
be  used  as  a  starting  motor.    An  induction  motor  is,  however, 
more  often  used. 

This  method  requires  the  use  of  synchronizing  apparatus  and 
the  installation  of  a  starting  motor,  but  causes  minimum  distur- 
bance in  the  alternating-current  system  during  the  starting 
period. 

(b)  As  an  induction  motor .f  —  If  the  field  circuit  of  a  synchro- 
nous motor  is  opened  and  an  alternating  current  supplied  to  the 
armature,  the  changing  magnetism  set  up  by  the  armature  currents 
induces  currents  in  the  field  pole  shoes.     These  currents,  reacting 
with  the  magnetism  set  up  by  the  armature  winding,  produce  a 
small  torque,  if  the  motor  is  polyphase,  and  the  unloaded  motor 
starts  without  the   use   of   auxiliary   devices.     When   the   motor 
attains  approximately  synchronous  speed,  which  is  indicated  by 
a  violent   swinging  of  the  pointer  of   an   ammeter   connected  in 
the  alternating-current  supply  line,  the  field  circuit  is  closed  and 
the  motor  pulls  into  step  with  the  supply  circuit. 

The  starting  torque  of  a  synchronous  motor,  when  started  in 
this  way,  is  increased  by  means  of  a  "squirrel  cage"  field  structure 
similar  to  that  shown  in  Fig.  i28b.  This  construction  also  tends  to 
prevent  " hunting,"  as  explained  in  Section  8. 

*  See  Chapter  8,  Section  17. 

.  f  This  method  of  starting  a  synchronous  motor  depends  on  the  fundamental  prin- 
ciples underlying  the  action  of  the  induction  motor,  and  will  not  be  understood  until 
those  principles  have  been  studied.  The  statements  made  here  are  to  be  taken  simply 
as  a  mechanical  process  by  which  the  synchronous  motor  may  be  started. 


THE   SYNCHRONOUS  MOTOR 


153 


Because  of  the  low  power  factor  and  the  large  current  required 
during  the  starting  period,  this  method  of  starting  may  cause  un- 
desirable disturbances  in  the  system  from  which  it  is  supplied. 
These  disturbances  are  particularly  objectionable  when  a  single 
motor  forms  a  considerable  portion  of  the  total  load  on  the  supply 
system  and  is  stopped  and  started  frequently,  or  when  motors  are 
operated  in  parallel  with  incandescent  lamps. 

5.  Stability.  —  An  engine  or  motor  is  in  stable  equilibrium  as 
long  as  an  increased  load  automatically  produces  a  corresponding 
intake  of  power.  As  noted  in  Section  3,  an  increased  load  on  a 
synchronous  motor  causes  the  angle  between  the  applied  and  the 
counter-electromotive  forces  to  increase.  Assuming  the  applied 


FIG.  131. 

voltage  to  be  constant,  the  operation  of  the  motor  is  stable  as  long 
as  the  product  of  the  current  and  the  power  factor  increases;  but 
it  is  evident  that,  as  the  angle  between  the  two  electromotive 
forces  increases,  a  point  is  finally  reached  where  the  power  factor 
decreases  faster  than  the  current  increases.  Beyond  this  point  the 
motor  is  in  unstable  equilibrium,  drops  out  of  step,  and  stops. 

The  above  statement  is  illustrated  by  Fig.  131,  in  which  OA  is 
the  vector  of  applied  electromotive  force,  and  OB  the  vector  of 
counter-electromotive  force.  OC  is  the  vector  of  the  resultant  elec- 
tromotive force,  and  OD  the  vector  of  current  flowing  in  the  arma- 
ture circuit.  The  impedance  of  the  circuit  is  approximately  con- 


154  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

stant,  and  OD  is  proportional  to  OC.  The  power  input  to  the 
motor  is  proportional  to  the  projection  of  the  current  vector  on 
the  vector  of  applied  electromotive  force,  i.e.,  to  the  product  of 
the  current  vector  and  the  cosine  of  angle  AOD. 

For  the  relation  of  the  vectors  shown  in  Fig.  13  la,  the  power  in- 
put to  the  motor  is  proportional  to  OE.  As  the  load  on  the  motor 
increases,  the  angle  0  increases  until  the  vector  of  the  counter- 
electromotive  force  reaches  the  position  indicated  in  Fig.  13  ic. 
If  the  angle  6  increases  still  further,  the  projection  OE  decreases 
and  the  power  input  to  the  motor  decreases  correspondingly,  as 
indicated  in  Fig.  13  id. 

Therefore,  if  a  synchronous  motor  is  loaded  beyond  a  certain 
point,  it  cannot  develop  sufficient  torque  to  carry  the  load,  and  will 
stop.  In  the  commercial  motor,  the  armature  current  usually  be- 
comes excessive  and  overheats  the  motor  before  it  reaches  the 
point  of  instability. 

6.  Maximum  load.  —  That  the  maximum  intake  of  a  synchro- 
nous motor  for  any  given  counter-electromotive  force  (field  ex- 
citation)  occurs  when  6  —  (3  =  180  degrees  is  evident  from   the 
following  considerations:    Referring  to  Fig.  1310  let 

6  -  0  =  180°.  (3) 

If  the  angle  6  is  either  increased  or  decreased,  both  the  projection 
of  the  current  vector  on  the  vector  of  the  electromotive  force  and 
the  input  to  the  motor  are  decreased. 

7.  Efficiency.  —  The  efficiency  of  a  synchronous  motor  is  deter- 
mined :  (a)  by  a  brake  test,  (b)  from  the  losses. 

(a)  Brake  test.  —  Measure  the  input  and  the  output,  and  compute 
the  efficiency  from  the  equation 

~  .  output  X  100  t  , 

Per  cent  efficiency  =  — ~ —        —  (4) 

input 

(b)  From  the  losses.  —  The  losses  in  a  synchronous  motor  are  the 
same  as  those  of  an  alternating-current  generator  and  are  deter- 
mined in  the  same  way.* 

8.  Hunting.  —  The   phenomenon   known   as   "  hunting "   in   a 
synchronous  motor,  or  a  rotary  converter,  consists  of  a  periodical 

*  See  Chapter  8,  Section  15. 


/THE   SYNCHRONOUS  MOTOR  155 

variation  of  the  speed  of  the  rotating  parts,  the  speed  being  alter- 
nately too  fast  or  too  slow,  and  is  indicated  by  the  swinging  of 
the  ammeter  pointer,  and  by  a  humming  noise  peculiar  to  this 
condition. 

Hunting  may  be  caused  by:  (a)  a  sudden  change  in  the  load,  (b) 
irregularities  in  the  speed  of  the  prime  mover  driving  the  generator 
from  which  the  motor  is  supplied,  (c)  faulty  design  of  the  motor. 
The  inertia  of  the  rotating  parts  of  the  motor  tends  to  damp  out 
any  oscillations  that  may  be  set  up,  the  addition  of  a  heavy  fly- 
wheel adding  to  this  damping  effect.  Hunting  is  often  guarded 
against,  in  the  design  of  a  synchronous  motor,  by  wedging  bars  of 
copper  between  the  tips  of  the  pole  shoes,  or  by  a  " squirrel  cage" 
structure  similar  to  that  shown  in  Fig.  128. 

As  long  as  the  angular  velocity  of  the  rotating  parts  of  the  motor 
is  constant,  the  copper  bars  have  no  effect,  but  when  hunting 
takes  place  the  oscillatory  motion  causes  a  relative  movement  of 
the  bars  and  the  flux  set  up  by  the  armature  winding.  This  move- 
ment induces  an  electromotive  force,  and  currents  flow  in  the  bars. 
The  reaction  between  the  currents  in  the  bars  and  the  flux  set 
up  by  the  armature  winding  tends  to  damp  out  the  oscillations, 
hence  the  name  "magnetic  dampers"  which  is  sometimes  applied  to 
this  construction.* 

9.  Phase  characteristic.  —  The  phase  characteristic,  commonly 
called  V-curve,  of  a  synchronous  motor  shows  the  relations  between 
the  armature  current  and  the  field  excitation,  the  load  remaining 
constant.  Referring  to  Fig.  132,  let 

OA  be  the  vector  of  the  applied  electromotive  force, 

OB  be  the  vector  of  the  counter-electromotive  force, 

OC  be  the  vector  of  the  resultant  electromotive  force, 

OD  be  the  vector  of  the  armature  current, 

ft  be  the  angle  between  the  current  and  the  resultant  electromo- 
tive force, 

<£  be  the  power  factor  angle, 

7  cos  <j>  be  constant,  i.e.,  the  locus  of  the  current  vector  is  a 
straight  line  perpendicular  to  the  vector  of  the  applied  electro- 
motive force. 

*  The  action  of  copper  bars  in  the  prevention  of  hunting  will  be  more  apparent 
after  the  induction  motor  has  been  studied. 


156 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


When  <f>  =  o  (current  and  applied  electromotive  force  in  phase), 
the  relations  in  the  circuit  are  represented  in  Fig.  1320;    when 

<f>  =  30  degrees  (lagging),  the  rela- 
tions in  the  circuit  are  represented 
in  Fig.  i32a;  and  when  $  =  30 
degrees  (leading),  the  relations  in 
the  circuit  are  represented  in  Fig. 

I32C. 

A  comparison  of  the  counter- 
electromotive  forces  in  Fig.  132, 
shows  that  the  power  factor  of  a 
synchronous  motor  is  dependent 
on  its  counter-electromotive  force, 
i.e.,  on  its  field  excitation.  The 
power  factor  of  the  motor  is,  therefore,  changed  by  changing  its 
field  excitation. 

Fig.  133  shows  phase  characteristics  for  different  loads.  For  a  con- 
stant load,  the  power  factor  of  the  motor  is  computed  as  the  ratio  of 
the  minimum  armature 
current  to  the  armature 
current  at  the  given  or 
required  field  excitation. 
Because  of  distortion  of 
the  wave  shape,  unity 
power  factor  is  seldom 
or  never  attained. 

10.  The  circle  dia- 
gram. —  In  any  alter- 
nating -  current  circuit 
having  a  constant 
applied  voltage,  a  COTi- 


Maximum 
\potver  factor 


L  eadinq 


FIELD  AMPERES 
FIG.  133.     Phase  Characteristics. 


stant   reactance   and   a 

variable    resistance,     it 

may  be  shown  that  the 

locus    of     the    current 

vector  is  a  semicircle.     The  phenomena  of  the  synchronous  motor 

may,  therefore,  be  represented  by  means  of  a  circle  diagram,  when  it 

is  assumed  that  the  synchronous  reactance  of  the  motor  is  constant.* 

*  This  assumption  simplifies  the  treatment,  and  the  error  introduced  is  not  great. 


THE   SYNCHRONOUS  MOTOR  157 

Lay  off  AO  and  OD  (Fig.  134)  proportional  to  the  applied  elec- 
tromotive force,  making  the  angle  AOD  such  that 

,  (5) 


when  Xa  =  the  synchronous  reactance  of  the  motor  armature, 
Ra  =  the  resistance  of  the  motor  armature. 

With  D  as  a  center  and  a  radius  proportional  to  the  counter- 
electromotive  force  at  any  given  or  required  field  excitation,  draw 
a  semicircle.      The  semicircle  is  the  locus 
of  the  current  vector  07,  and   the  react- 
ance    drop    OC,   the    resistance   drop    BC, 
and   the   counter   electromotive   force   AB 
may  be  drawn  in  their  correct  phase  rela- 
tions and  magnitudes  as  indicated  in  the  ^^    circle  Diagram. 
figure. 

ii.  The  synchronous  phase-modifier.  —  As  shown  in  Fig.  I32C, 
a  synchronous  motor,  when  over  excited,  takes  a  leading  current 
from  the  supply  system.  The  wattless  component  of  a  leading 
current  opposes  and  neutralizes  an  equal  wattless  component  in  a 
transmission  line  or  a  distributing  system  *  due  to  a  parallel  induc- 
tive load  such  as  an  induction  motor.  When  used  for  the  purpose 
of  improving  the  power  factor  of  a  circuit,  a  synchronous  motor  is 
termed  a  "synchronous  phase-modifier,"  although  it  may  deliver 
mechanical  power  at  the  same  time. 

Example.  —  Two  ioo-h.p.,  440  volt  induction  motors  operate 
at  85  per  cent  power  factor  and  87  per  cent  efficiency.  Find  the 
saving  in  transmission  line  copper  when  one  of  the  induction  mo- 
tors is  replaced  by  a  synchronous  motor,  the  efficiency  of  which  is 
85  per  cent  when  over  excited  so  as  to  compensate  for  the  lagging 
current  of  the  other  induction  motor.  Also  find  the  rated  output 
of  the  synchronous  motor,  and  its  power  factor. 

Solution. 

T  =  200  X  746 

"  0.85  X  0.87  X  440 

=  465  amperes  for  two  induction  motors, 

*  See  Chapter  i,  Section  38. 


158  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

,  _   100  X  746         100  X  746 
0.87  X  440       0.85  X  440 
=  194  +  199 
=  393   amperes  for  one  induction  motor 

and  one  synchronous  motor, 
R'  (465)2  =  R"  (393)2 


i.e.,  the  weight  of  copper  is  approximately  40  per  cent  greater  for 
two  induction  motors  than  for  one  induction  motor  and  one  syn- 
chronous motor. 

P.P.  of  snchronous  motor  = 


(233   X  0.52) 


234 
=  0.85. 

Rating  of  synchronous  motor    =  - 

0.85 

=  119  h.p.  (use  125  h.p.). 

12.  Load  division  of  alternators  in  parallel.  —  Let  A  and  B  be 

two  similar  alternating-current  generators  operating  in  parallel, 
the  field  excitation,  terminal  voltage  and  armature  current  being 
the  same  in  each  machine. 

If  the  power  input  to  the  prime  mover  driving  alternator  B  is 
reduced,  the  load  on  the  alternator  tends  to  stop  its  rotating  parts, 
the  angle  between  the  electromotive  forces  becomes  greater  than 
1  80  degrees,  and  a  current  flows  between  the  armatures,  i.e.,  alter- 
nator B  acts  simultaneously  as  an  alternating-current  generator 
and  as  a  synchronous  motor,  the  current  for  its  motor  action  being 
received  from  alternator  A.  Therefore,  the  load  division  of  alter- 
nating-current generators  operating  in  parallel  is  changed  by  chang- 
ing the  power  input  to  their  prime  movers. 

Since  the  input  to  a  prime  mover,  e.g.,  a  steam  engine,  is  constant 
at  a  given  speed,  changing  the  field  excitation  of  alternator  B  does 
not  change  its  output,  although  it  causes  a  current  to  flow  between 
the  armatures.  Since  the  resultant  electromotive  force  is  in  phase 
with  that  induced  in  alternator  A  and  the  angle  0  is  nearly  90  de- 
grees, it  is  evident  that  this  current  is  wattless  except  for  the  small 


THE   SYNCHRONOUS  MOTOR  159 

power  component  due  to  the  resistance  of  the  circuit,  that  it  leads 
the  electromotive  force  of  alternator  B  and  lags  behind  the  electro- 
motive force  of  alternator  A.  This  quadrature  current  can  pro- 
duce no  torque  in  either  machine  and  the  load  distribution  of 
alternators  operating  in  parallel  is,  therefore,  not  disturbed  by  a 
change  in  field  excitation. 

Since  the  current  flowing  between  the  armatures  leads  the  elec- 
tromotive force  of  alternator  B  and  lags  behind  the  electromotive 
force  of  alternator  A,  its  effect  is  to  equalize  the  voltages  induced 
in  the  armatures  by  reducing  the  flux  set  up  by  one  field  winding 
and  increasing  that  set  up  by  the  other. 


CHAPTER  IX  — PROBLEMS 

1.  Show,  by  means  of  the  circle  diagram,  that  the  power  factor  of  a  synchro- 
nous motor  is  changed  by  changing  its  field  excitation,  the  load  remaining 
constant. 

2.  The  synchronous  impedance  of  an  alternator  is  1.5  ohms.    When  oper- 
ated as  a  synchronous  motor  from  22o-volt  mains  and  without  load,  the  arma- 
ture current  is  10  amperes,  and  the  power  factor  is  0.866  leading.    The  angle 
ft  —  85  degrees.     Find  the  counter-electromotive  force  of  the  motor. 

3.  Find  the  counter-electromotive  force  of  the  motor  in  Problem  2  when  it 
power  factor  is  unity. 

4.  Find  the  power  factor  of  the  motor  in  Problem  2  when  the  applied  volt- 
age is  reduced  10  per  cent. 

5.  Calculate  the  armature  resistance  of  the  motor  in  Problem  2. 

6.  Draw  the  vector  diagram  for  the  three-phase  alternator  of  Problem  n, 
Chapter  8,  when  operated  as  a  motor  at  unity  power  factor,  and  rated  armature 
current. 

7.  From  the  diagram  obtained  in  Problem  6,  calculate  the  torque  developed 
in  the  armature. 

8.  A  3-phase,   23oo-volt,   5oo-kv-a.  synchronous  motor  is  overexcited  so 
that  its  power  factor  is  0.6  leading.      Calculate  the  capacitance  to  which  it 
is  equivalent  when  operated  from  6o-cycle  mains,  and  full-load  (rated)  current 
flows  in  the  armature  circuit. 

9.  A  3-phase,  22oo-volt  induction  motor  having  a  lagging  power  factor  of 
0.85  takes  zoo  amperes  (line),  and  is  connected  in  parallel  with  an  over- 
excited synchronous  motor.    When  the  power  factor  of  the  system  is  unity  the 
current  (line)  supplied  to  the  synchronous  motor  =125  amperes.     Find:  (a) 
the  power  factor  of  the  synchronous  motor,  (b)  the  line  current,  (c)  the  input 
(watts)  to  each  motor.     Draw  a  diagram  representing  the  electromotive  fc 
and  the  currents  in  the  system. 


160  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

10.  Plot  the  circle  diagram  for  the  alternator  in  Problem  n,  Chapter  8, 
when  operated  as  a  synchronous  motor,  and  from  this  diagram  construct  the 
phase  characteristic  (V-curve)  for  an  input  of:  (a)  500  kw.,  (6)  100  kw. 

n.   Same  as  Problem  10  except  the  resistance  of  the  armature  is  2  ohms. 

12.  Same  as  Problem  10  except  the  resistance  of  the  armature  is  0.5  ohm. 

13.  The  alternator  in  Problem  2  is  used  as  a  phase  modifier.    The  armature 
current  is  100  amperes  and  the  power  factor  o.i.     Find:  (a}  the  power  input 
to  the  armature,  (6)  the  wattless  component  of  lagging  current  that  it  will 
neutralize. 

14.  The  alternator  hi  Problem  2  is  driven  as  a  synchronous  motor  from  220 
volt  mains.     The  counter-electromotive  force  is  220  volts.     Find:  (a)  the  cur- 
rent input  to  the  motor  at  no  load,  (6)  the  power  factor  at  which  the  motor 
operates,  (c)  the  phase  angle  between  the  applied  and  the  counter-electromotive 
forces. 


CHAPTER  X 
CURRENT-RECTIFYING   APPARATUS 

SINCE  electrical  energy  is  most  economically  and  satisfactorily 
transmitted  by  means  of  alternating  currents,  and  many  commercial 
applications  require  unidirectional  currents,  it  is  necessary  to  pro- 
vide means  for  the  rectification  or  conversion  of  alternating  cur- 
rents. Alternating  currents  may  be  made  unidirectional  by:  (a)  a 
mechanically-driven  commutator,  (b)  a  motor-generator,  (c)  the 
mercury  arc  rectifier,  (d)  the  synchronous  converter. 

(a)  The  Commutator 

1.  A  commutator  driven  by  a  small  synchronous  motor  would 
appear  to  be  a  cheap  and  an  efficient  means  for  the  rectification  of 
alternating  currents.     This  method  is,  however,  of  little  commercial 
importance  because  of  its  limited  capacity,  excessive  sparking,  with 
a  consequent  destruction  of  the  commutator,  taking  place  whenever 
either  the  current  or  the  electromotive  force  increases  above  very 
limited  values. 

(b)  The  Motor-Generator 

2.  The  motor-generator  consists  of  a  motor,  either  synchronous 
or  induction,  which  drives  a  continuous-current  generator.     The 
apparatus  differs  in  no  way  from  that  described  elsewhere  in  this 
volume.     Motor-generators   are  in   extensive  use   where   a   wide 
variation  of  the  continuous  voltage  is  required.     Because  it  con- 
sists of  two  machines,  the  first  cost  of  a  motor-generator  is  high, 
and  the  operating  efficiency  low. 

(c)  The  Mercury  Arc  Rectifier 

3.  Construction.  —  The  mercury  arc  rectifier  consists  of  a  highly 
exhausted  glass  tube  having  two  or  more  iron  or  graphite  terminals 
to  which  the  alternating-current  supply  circuit  is  connected,  and  a 
mercury  terminal  to  which  the  load  circuit  is  connected.     Fig.  135. 

161 


162 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


It  is  also  provided  with  an  auxiliary  mercury  terminal  which  is 
used  for  starting  purposes  only.  The  circuits  for  a  single-phase 
rectifier  are  shown  in  Fig.  i36a,  and  those  for 
a  three-phase  rectifier  in  Fig.  1360. 

The  mercury  arc  rectifier  depends,  for  its 
operation,  on  the  fact  that  an  electric  current 
may  flow  from  iron  or  graphite  terminals  to 
mercury,  but  may  not  flow  in  the  reverse  direc- 
tion. The  flow  of  current  in  the  rectifier  circuit 
is  similar  to  that  of  water  in  the  hydraulic  sys- 
tem shown  in  Fig.  137,  where  check  valves  VV 


FIG.  135.    Mercury  Arc    permit  water  to  flow  only  in  the  direction  indi- 
RectifierTube.  West-    cate(j  ty  the  arrows>     in  the  single-phase  rec- 
tifier, current  flows  from  terminal  A   to  the 
mercury  during  one-half  cycle,  and  from  terminal  B  to  the  mercury 
during  the  other  half  cycle.     The  current  in  the  load  circuit  is, 
therefore,  a  unidirectional  (pulsating)  current. 


FIG.  136.     Wiring  Diagrams  for'Mercury    FIG.  137.     Hydraulic  Analogy  for  Mer- 
Arc  Rectifier.  cury  Arc  Rectifier. 

4.  Starting. — The  mercury  arc  rectifier  cannot  be  started  simply 
by  connecting  the  iron,  or  graphite,  terminals  to  an  alternating- 
current  supply  circuit,  because  of  the  high  initial  resistance  of  the 
tube.  The  auxiliary  mercury  terminal  is,  therefore,  necessary  to 
break  down  this  high  resistance.  When  the  starting  switch  is 
closed,  and  the  bulb  tilted  slightly  so  that  the  mercury  terminals 
are  connected  by  the  liquid  mercury  in  the  tube,  an  electric  current 
flows  through  the  mercury  and  fills  the  tube  with  vapor.  The 
mercury  vapor  reduces  the  resistance  of  the  main  circuit  so  that 


CURRENT-RECTIFYING  APPARATUS 


163 


current  may  flow  between  the  iron  terminals  and  the  mercury. 
The  tube  may  now  be  allowed  to  return  to  its  normal  position  and 
the  starting  switch  opened,  the  load  circuit  having  been  closed 
through  a  suitable  resistance,  since  the  arc  will  " break"  and  the 
rectifier  cease  to  operate  if  the  current  falls  below  a  certain  min- 
imum value. 

5.  Operation.  —  Because  the  arc  "  breaks "  when  the  load 
current  falls  below  a  certain  minimum  value,  which  varies  from 
two  to  four  amperes,  an  inductance  must  be  placed  in  the  load 
circuit  so  that  the  decay  of  the  current  established  during  one- 
half  cycle  is  delayed  until  that  from  the  other  half  cycle  can  be 
established  *  making  the  form  of 
the  wave  as  shown  in  Fig.  138. 
Without  this  inductance,  which  may 
be  the  secondary  of  the  transformer 
supplying  the  rectifier,  the  current 
decreases  to  zero  at  the  end  of  each 
half  cycle  and  the  operation  of  the 
arc  is  interrupted.  Because  of  the 
overlapping  of  the  waves  in  the  dif- 
ferent phases  of  a  polyphase  system,  no  inductance  is  required  in 
the  load  circuit  of  a  rectifier  using  polyphase  currents. 

The  voltage  Edc  of  the  load  circuit  bears  a  constant  ratio  to  the 
voltage  Eac  of  the  supply  circuit. 


7,; 


Zeio  L 


>-£ 


Current  Wore  in 

I 

PtoiHye  Electrode 
Current  Wavem 

H 
fbstrivt  Electrode 

Wave  of  Rectified 

m 

Current 

Wave  of  Impressed 
IV 


FIG.  138.     Wave  Forms  Mercury 
Arc  Rectifier. 


Ed,    = Eac  —  k, 

IT 


d) 


when  k  is  the  counter-electromotive  force  of  the  tube  which  varies, 
in  different  tubes,  from  thirteen  to  fifteen  volts. 

Because  of  the  high  temperatures  at  which  they  operate,  the 
tubes  of  commercial  rectifiers  are  usually  immersed  in  oil  for  the 
better  dissipation  of  the  heat. 

The  mercury  arc  rectifier  delivers  constant  electromotive  force 
or  constant  current,  as  the  transformer  T  is  constant  potential  or 
constant  current. 

6.  Efficiency.  —  From  equation  (i),  it  is  evident  that  from 
thirteen  to  fifteen  volts  are  lost  in  the  rectifier  tube,  whatever 
the  applied  voltage  may  be.  Therefore,  the  efficiency  is  greater 

*  This  statement  applies  to  single-phase  rectifiers  only. 


164  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

at  high  voltages  than  at  low,  but  is  constant  at  all  loads.  At  low 
voltages,  such  as  are  commonly  used  for  motors,  the  efficiencies  of 
mercury  arc  rectifiers  do  not  compare  favorably  with  other  types 
of  rectifying  apparatus. 

7.  Limitations  and  use.  —  Because  of  its  low  efficiency  at  the 
voltages  commonly  used  for  continuous-current  apparatus,  and  its 
poor  power  factor,  the  use  of  the  mercury  arc  rectifier  is  restricted, 
at  the  present  time,  to  special  purposes.     It  is  largely  used  for 
supplying  unidirectional  current  to  arc  lamp  systems  where  the 
high  voltage  and  the  small  current  required  make  it  very  satis- 
factory when  operated  in  connection  with  constant-current  trans- 
formers.    It   is   also   used,   in   small  units,   for  charging  storage 
batteries. 

(d)  The  Synchronous  Converter 

8.  Construction.  —  The  synchronous,  or  rotary,  converter  does 
not  differ  essentially  from  a  shunt  or  compound  (continuous-cur- 
rent) dynamo  with  the  addition  of  symmetrical  connections  from 

the  armature  winding  to  two  or  more 
continuous  rings  mounted  on  the  arma- 
ture shaft  in  the  same  manner  as 
for  an  alternating-current  generator 
with  rotating  armature.  Fig.  139. 
Practically,  the  converter  which  has 
more  than  two  rings  differs  from  the 
continuous  -  current  machine  of  the 
same  rating  by  having  a  much  larger 
FIG.  i39.  Synchronous  Converter  commutator,  and  smaller  yoke  and 

Armature.     Westinghouse.  ,,   ,  ,        , 

field  poles. 

9.  Operation.  —  A  synchronous  converter  may  be  operated  as: 

(a)  A  continuous-current  motor. 

(b)  A  continuous-current  generator. 

(c)  A  synchronous  motor. 

(d)  An  alternating-current  generator. 

(e)  A  double-current  generator,  i.e.,  simultaneously  as  a  contin- 
uous-current generator  and  an  alternating-current  generator. 

(/)  A  converter,  i.e.,  simultaneously  as  a  synchronous  motor 
and  a  continuous-current  generator.  When  so  operated,  the  arma- 
ture must  rotate  at  synchronous  speed,  and  the  speed  of  any  given 


CURRENT-RECTIFYING  APPARATUS  165 

converter  is  determined  by  the  frequency  of  the  supply  circuit. 
The  converter  exhibits  the  same  phase  characteristic,  power  factor 
and  line-current  changing  with  the  field  excitation,  as  the  synchro- 
nous motor. 

(g)  An  inverted  converter,  i.e.,  simultaneously  as  a  continuous 
current  motor  and  an  alternating-current  generator.  When  so 
operated,  the  speed  and,  consequently,  the  frequency  of  the  alter- 
nating current,  varies  with  the  field  excitation,  unless  operated  in 
parallel  with  other  synchronous  apparatus  which  controls  the  fre- 
quency of  the  circuit.  Since  the  armature  reaction  of  an  alternat- 
ing-current generator,  the  load  circuit  of  which  is  inductive,  tends 
to  demagnetize  the  field,  the  speed  of  an  inverted  synchronous  con- 
verter supplying  such  a  circuit  may  become  excessive,  and  means 
should  be  provided  for  preventing  these  high  speeds  which  may 
wreck  the  machine.  Two  methods  are  employed  to  limit  the 
speed  of  an  inverted  converter  to  safe  values:  (i)  separate  excita- 
tion, (2)  speed-limit  switches. 

(1)  Separate    excitation.  —  If    an    inverted    converter,    instead 
of  being  excited  from  the  supply  mains,  has  its  field  windings 
supplied  from  a  separate  exciter  direct  connected,  or  belted,  to 
the  armature   shaft,   an   increase   or   decrease  in   the   speed   of 
the    armature   causes   a    corresponding   increase   or   decrease  in 
the  excitation  of  the  converter,   and    tends  to  keep  the   speed 
constant. 

(2)  Speed-limit  switches.  —  A  speed-limit  switch  is  attached  to  the 
shaft  of  the  converter,  and  operates  circuit  breakers  in  the  con- 
tinuous-current supply  mains  when  the  speed  reaches  a  prede- 
termined value,  thus  stopping  the  machine. 

10.  Voltage  relations.  —  The  voltage  between  the  brushes  of  a 
continuous-current  generator  is  the  algebraic  sum  (constant)  of  the 
instantaneous  electromotive  forces  induced  in  the  different  coils  con- 
nected in  series  between  the  brushes;  the  maximum  alternating 
voltage  induced  in  the  same  coils  is  equal  to  the  geometric  sum 
of  the  maximum  electromotive  forces  induced  in  the  separate 
coils.  But  the  geometric  sum  of  the  maximum  alternating-electro- 
motive forces  induced  in  the  distributed  windings  of  a  synchronous 
converter,  between  adjacent  continuous-current  brushes,  is  the  alge- 
braic sum  of  the  instantaneous  electromotive  forces.  Therefore, 
the  continuous  voltage  of  a  synchronous  converter  is  equal  to  the 


i66 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


maximum  alternating-electromotive  force  induced  in  the  coils  con: 
nected  in  series  between  adjacent  continuous-current  brushes. 

Let  an  armature  have  twelve 
coils  symmetrically  placed,  as  indi- 
cated in  Fig.  140.  The  maximum 
alternating  electromotive  forces 
induced  in  the  coils,  and  their 
phase  relations,  are  represented  by 
a  twelve-sided  polygon,  and  the 
maximum  alternating  voltage  be- 
tween continuous-current  brushes 
is  equal  to  the  geometric  sum  of 
the  maximum  electromotive  forces 
induced  in  any  six  coils,  i.e.,  to 
the  diameter  of  the  circumscribed  circle.  Fig.  141.  The  sum  of  the 
instantaneous  electromotive  forces  induced  in  the  coils  between 
continuous-current  brushes  is  also  the  diameter  of  the  circumscribed 
circle.  If  the  armature  winding  is  connected  to  two  rings,  the  maxi- 
mum alternating  voltage  between  rings  is  equal  to  the  voltage  be- 
tween the  continuous-current  brushes, 
and  the  effective  alternating  voltage  is 
equal  to  the  continuous  voltage  divided 
by  the  square  root  of  2.  Fig.  141. 


FIG.  140.     Single-phase  Rotary 
Converter. 


Six  King 


=  £« 


(2) 

(3) 


FourR'mq 


< 


Similarly,  if  the  armature  is  provided 
with  three  rings,  the  maximum  alternating  voltage  between  rings  is 
the  geometric  sum  of  the  maximum  voltages  induced  in  four  coils,  or 
the  side  of  the  inscribed  triangle;  if  four  rings,  the  geometric  sum  of 
the  maximum  voltages  induced  in  three  coils,  or  the  side  of  the  in- 
scribed square;  if  six  rings,  the  geometric  sum  of  the  maximum 
voltages  induced  in  two  coils,  or  the  side  of  the  inscribed  hexagon. 

Therefore,  the  effective  alternating  voltage  between  adjacent 
rings  of  an  w-ring  synchronous  converter  required  to  give  a  speci- 
fied continuous  voltage  Ecc  is 

1 80° 


sin 


Eac   == 


(4) 


CURRENT-RECTIFYING  APPARATUS  167 

It  will  be  remembered  that  the  phase  voltage  of  a  four-ring  (two- 
phase)  converter  is  the  voltage  between  alternate  rings.  Hence, 
the  phase  voltage  of  a  four-ring  converter  is  equal  to  the  phase 
voltage  of  a  two-ring  converter,  the  continuous  voltage  remaining 
constant. 

TABLE  IV 

Eac  for  a  2-ring  converter  =  0.707  times  the  continuous  voltage 

3-  =0.612 

4-  =0.500 
6-  =0.354 

From  the  above  considerations  it  is  evident  that  the  ratio  be- 
tween the  alternating  and  the  continuous  voltages  of  a  synchronous 
converter  is  fixed,  i.e.,  the  voltage  at  the  continuous-current  brushes 
can  be  changed  only  by  changing  the  voltage  between  the  alter- 
nating-current rings.  The  theoretical  ratios  developed  above  are 
obtained,  within  very  narrow  limits,  in  the  commercial  converter. 
The  practical  considerations  which  affect  the  ratios  are:  (i)  the 
resistance  of  the  armature  windings,  (2)  the  shape  of  the  alternat- 
ing electromotive-force  wave. 

(1)  Armature   resistance.  —  The   theoretical   ratios   of   voltages 
given  above  are  changed  by  virtue  of  the  resistance  of  the  armature 
windings  but  this  resistance  is  always  small,  and  the  ratios  are  never 
seriously  affected. 

(2)  Wave  shape.  —  When  the  electromotive  force  *wave  is  not 
harmonic,   the   effective   value   is    not    equal    to    the    maximum 
divided  by  the  square  root  of  2.     For  example,  for, a  flat-topped 
wave  the  effective  value  is  greater  than  for  a  sine  wave  having 
the  same  maximum,  while  for  a  peaked  wave,  the  effective  value  is 
less.     In  a  well-designed  converter,  the  wave  shape  is  not  seriously 
distorted. 

ii.  Current  ratios.  —  The  ratio  of  the  current  in  the  alternating- 
current  circuit  to  that  in  the  continuous-current  circuit  of  a  con- 
verter is  easily  calculated  when  the  losses  in  the  converter  are 
neglected,  i.e.,  when  the  output  is  considered  equal  to  the  input. 

In  an  w-ring  two-pole  armature,  the  alternating  voltage  between 
the  neutral  and  any  ring  is 

£»-•%•  (5) 

2  V  2 


168  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

and  *^  =  EJCCJ  (6) 

2  V  2 

when  the  power  factor  of  the  converter  is  unity.     Therefore,  the 
line  current 


and  the  current  hi  the  armature  conductors 

,    ,  _        V2lcc 
IOC    '        ~ 


wsin 


n 


12.  Starting.  —  The  preferable  way  to  start  a  synchronous  con- 
verter is  as  a  continuous-current  motor,  or  by  means  of  a  small 
starting  motor,  after  which  the  alternating-current  end  is  synchro- 
nized *  with  the  system  from  which  the  converter  is  to  be  supplied. 
When  no  continuous-current  energy  is  available,  the  converter  may 
be  started  as  an  induction  motor,  as  described  for  the  synchronous 
motor.f    The  same  disturbances  are  caused  in  the  alternating-cur- 
rent system  as  with  the  synchronous  motor,  and  the  polarity  of  the 
continuous-current  brushes  must  be  determined  before  connection  is 
made  to  a  storage  battery,  or  other  apparatus  through  which  the 
current  flows  in  a  given  direction.    The  polarity  of  the  continuous- 
current  brushes  is  fixed  by  the  last  pulsation  of  the  alternating  cur- 
rent as  the  armature  pulls  into  step  with  the  supply  circuit. 

13.  Compounding.  —  Since    an    increased    field    excitation  of 
a  converter,   as  of  a  synchronous  motor,  is  neutralized  by  an 
increased  wattless  component  of  armature  current,  compounding 
of  a  synchronous  converter  is  not  possible,  in  the  sense  that  the  term 
is  used  in  connection  with  continuous-current  dynamos.     The  only 
way  to  increase  the  continuous  voltage  of  a  converter  is  to  increase 
the  alternating  voltage  applied  at  the  rings.     A  series   or  com- 
pound field  winding  may  be  made  to  increase  the  continuous 
voltage,  provided  the  alternating-supply  circuit  is  inductive. 

If  the  alternating-current  circuit  supplying  a  converter  is 
inductive,  the  inductive  drop  in  the  line  makes  the  voltage  at  the 
rings  less  than  it  would  be  if  the  circuit  were  non-inductive. 

*  See  Chapter  8,  Section  17. 
t  See  Chapter  9,  Section  4. 


CURRENT-RECTIFYING  APPARATUS 


169 


When  a  synchronous  motor  is  over  excited,  a  leading  current  flows 
in  the  alternating-current  system,*  and  the  circuit  is  equivalent  to 
a  series  circuit  containing  resistance,  inductance  and  capacitance. 
The  inductive  line  drop  is  thus  neutralized,  partially  or  completely, 
and  the  voltage  at  the  rings  of  a  converter,  and  consequently,  the 
continuous  voltage,  is  increased.  Because  its  average  power  factor 
is  inherently  low,  the  operation  of  a  compound-wound  converter  is 
unsatisfactory. 

14.  The  regulating  (or  split)  pole  converter.  —  For  automati- 
cally increasing  the  continuous-voltage  as  the  load  on  the  con- 
verter increases,  the  regulating  pole  (the  so-called  "split"  pole) 
converter  has  been  designed.  In  this  type  of  converter,  the  field 
pole  is  divided  into  two  parts,  each  part  has  its  own  winding,  and 
may  be  excited  to  any  required  degree  independently  of  the  other 
part.  If  the  regulating  pole  is  unexcited,  the  converter  operates  in 
the  usual  manner  and  the  continuous  voltage  of  a  two-ring  con- 
verter is  equal  to  the  maximum  in- 
duced alternating:electromotive  force. 

Let  Fig.  142  represent  a  two-ring 
converter  having  two  sets  of  field 
poles,  the  smaller  (regulating)  poles 
being  placed  midway  between  the 
larger  (main)  poles.  When  the  con- 
verter is  in  operation,  the  maximum  C. 
counter-electromotive  force  (OA ,  Fig. 
1 43  a)  induced  in  the  armature  wind- 


FIG.  142.    The  Regulating  Pole 
Converter. 


ings  is  equal,  neglecting  resistance  drop,  to  the  maximum  applied 
electromotive  force.  But  the  counter-electromotive  force  is  the 
resultant  of  two  quadrature  electromotive  forces,  as  shown  in  Fig. 


Cb) 


FIG.  143. 


1 43 a,  one  induced  in  the  winding  as  it  passes  under  the  main  poles, 
the  other  induced  in  the  winding  as  it  passes  under  the  regulating 
poles,  i.e.,  this  construction  is  electrically  equivalent  to  two  arma- 


*  See  Chapter  9,  Section  9. 


1 70  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

ture  windings  in  series,  the  induced  electromotive  forces  in-  which 
are  90  degrees  out  of  phase. 

The  continuous  voltage  of  a  synchronous  converter  is  the  alge- 
braic sum  of  the  instantaneous  electromotive  forces  induced  in  the 
armature  conductors  connected  in  series  between  adjacent  brushes 
(Section  10).  The  sum  of  the  instantaneous  electromotive  forces 
induced  in  the  armature  winding  in  cutting  the  flux  of  the  main 
pole  is  proportional  to  OB  (Fig.  143 a),  and  the  sum  of  the  instan- 
taneous electromotive  forces  induced  in  the  armature  winding  in 
cutting  the  flux  of  the  regulating  pole  is  proportional  to  OC  (Fig. 
i43a).  If  the  main  and  the  regulating  poles  between  brushes  are 
both  north  or  both  south,  the  continuous  voltage  is  proportional  to 
the  arithmetical  sum  of  the  lines  OB  and  OC;  if  the  main  and  the 
regulating  poles  between  brushes  are  not  of  the  same  polarity,  the 
continuous  voltage  is  proportional  to  the  arithmetical  difference  of 
the  lines  OB  and  OC. 

In  practice,  the  regulating  poles  are  not  placed  midway  between 
the  main  poles.  Under  this  construction,  the  phase  angle  between 
the  components  of  the  counter-electromotive  force  is  less  than  90 
degrees,  as  shown  in  Fig.  Msb. 

From  the  above  considerations,  it  is  evident  that  the  continuous 
voltage  of  a  regulating  pole  converter  is  changed  by  varying  the 
field  excitation.  Converters  of  this  type  are  in  practical  operation 
where  the  maximum  continuous  voltage  is  150  per  cent  of  the 
minimum  value,  the  alternating  voltage  remaining  constant. 

During  the  development  of  the  regulating  pole  converter,  fears 
were  expressed  that  its  alternating  electromotive-force  wave  would 
be  so  badly  distorted  as  to  make  its  use  undesirable.  These  fears 
have  been  proven  entirely  groundless. 

15.  The   Booster-Converter.  —  The  booster-converter  consists 
of  an  alternating-current  generator  and  a  synchronous  converter, 
the  armatures  of  which  are  connected  in  series.     By  means  of  the 
field  rhetostat  and  a  field  reversing  switch,  the  alternating  voltage 
supplied  to  the  converter  armature  may  be  either  increased   or 
decreased,  and  the  direct  voltage  varied  over  a  range  propor- 
tional to  twice  the  alternating  electromotive  force  induced  in  the 
generator  armature. 

1 6.  Hunting.  —  A  synchronous  converter,  because  of  the  light 
weight  of  its  rotating  parts,  is  more  susceptible  to  oscillatory  dis- 


CURRENT-RECTIFYING  APPARATUS  171 

turbances  than  is  a  synchronous  motor.  The  same  things  tend 
to  cause  hunting  in  a  converter  as  in  a  motor,  and  the  same 
methods  are  used  to  prevent  oscillations,  or  to  damp  them  out  when 
once  started. 

17.  Efficiency.  —  The  efficiency  of  a  synchronous  converter, 
which  is  usually  higher  than  that  of  either  a  motor  or  a  generator 
having  the  same  rating,  is  easily  determined  by  loading,  since  both 
input  and  output  are  electrical  quantities  which  are  easily  and 
accurately  measured. 

The  approximate  efficiency  may  be  calculated  from  the  rating, 
the  input  at  no  load,  measured  from  either  the  continuous-current 
or  the  alternating-current  end,  and  the  resistance  of  the  armature 
winding  as  measured  between  continuous-current  brushes. 


Per  cent  efficiency  =  (E  +  l  +  w>  fe) 

when  E  =  the  rated  continuous  voltage, 
Eb  =  brush-contact  resistance  drop.* 
/  =  the  continuous-  current  output  for  which  it  is  desired 

to  calculate  the  efficiency, 
W  =  the  no-load  input, 
Ra  =  the  resistance  of  the  armature  winding   as  measured 

between  continuous-current  brushes, 

k  =  the  ratio  of  the  copper  loss  in  the  armature  conductors 
to  the  armature  copper  loss  when  the  converter  is 
operated  as  a  continuous-current  generator  with  the 
same  output  (see  Section  18).  The  value  of  k  depends 
on  the  number  of  rings  with  which  the  converter  is 
provided  and  may  be  taken  from  Table  V. 

From  the  above,  it  is  evident  that  the  efficiency  of  a  converter 
increases  as  the  number  of  alternating-current  rings  is  increased. 
Converters  having  more  than  six  rings  are  seldom  used,  because 
the  increased  efficiency  is  not  sufficient  to  justify  the  increased  ex- 
pense and  complications. 

The  values  of  k  given  in  Table  V  assume  that  the  converter  is 
operated  at  unity  power  factor.  When  the  power  factor  is  less 
than  unity,  the  armature  copper  losses  are  increased  (see  Section 
19),  and  the  efficienty  is  decreased. 

*  See  Chapter  4,  Section  2. 


172 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

TABLE   V 


No.  rings 

k 

2 

i-39 

3 

0.56 

4 

0-37 

6 

o.  26 

8 

0.  21 

1 8.  Armature  reaction.  —  It  was  shown  in  Chapter  4,  Section  6, 
that  the  magnetomotive  force  set  up  by  the  current  in  the  armature 
of  a  two-pole  continuous-current  generator  is 


NI 


=  —  av.  cos 


NI 

—  ampere-turns. 

2  7T 


do) 


The  alternating  current  flowing  in  the  coils  of  an  w-ring  two-pole 
converter,  the  continuous-current  output  of  which  is  7,  is,  from 
equation  (8) , 

,    ,  V~2l  ,      . 

(12) 


and  the  ampere-turns  per  ring  are 

NIJ       -v/2  NI 


2  n 


9     .      TT 

2  n2  sin  - 
n 


Therefore,  the  maximum  magnetomotive  force  per  ring  is 


NI 


av.  cos 


.      TT 

sin- 
n 


n 


(14) 


TTH 


ampere- turns. 


When  the  current  and  the  electromotive  force  of  an  w-ring  arma 
ture  are  in  phase,  the  armature  magnetomotive  force  is  in  quadra 
ture  with  the  field  flux,*  and  is  equal  to 

*  See  Chapter  8,  Section  9. 


CURRENT-RECTIFYING  APPARATUS  173 

NI 

Mn  =  —  ampere- turns,*  (16) 

2  7T 

i.e.,  when  the  continuous-current  brushes  of  a  polyphase  converter 
are  set  midway  between  the  pole  tips,  the  armature  reaction  due 
to  the  continuous  current  is  equal  and  opposite  to  that  due  to  the 
power  component  of  the  alternating  current.  The  armature 
reaction  in  a  polyphase  converter  is,  therefore,  that  due  to  the 
current  required  to  supply  the  losses  in  the  converter,  and  to  the 
wattless  component  of  current  when  the  power  factor  is  less  than 
unity. 

From  equation  (15)  it  is  evident  that  the  maximum  armature 
magnetomotive  force  of  a  two-ringf  converter  is 

NI 

M2  =  —  ampere-turns,  (17) 

i.e,,  the  maximum  value  of  the  armature  magnetomotive  force  of 
a  two-ring  converter,  due  to  the  power  component  of  the  alternating 
current,  is  twice  that  due  to  the  continuous  current.  The  armature 
magnetomotive  force  of  a  two-ring  converter,  therefore,  varies 

NI  NI 

from  -f-  -    :   to ,  the  frequency  of  the  variation  being  twice 

2  7T  2  7T 

that  of  the  alternating-current  circuit.}; 

The  shifting  of  the  armature  flux  across  the  pole  faces  causes 
energy  losses  (hysteresis  and  eddy  currents),  and  tends  to  cause 
sparking  at  the  continuous-current  brushes.  Therefore,  commuta- 
tion in  a  two-ring  converter  is  inherently  poor,  and  is  not  improved 
by  an  angular  advance  of  the  brushes. 

19.  Armature  heating.  —  Since  a  synchronous  converter  acts 
simultaneously  as  a  continuous-current  generator  and  a  synchro- 
nous motor,  or  as  a  continuous-current  motor  and  an  alternating- 
current  generator,  the  currents  actually  flowing  in  the  armature 
conductors  must  be  resultants  of  these  two  actions.  The  instan- 
taneous current  flowing  in  an  armature  coil  of  a  converter  is,  then, 
the  algebraic  sum  of  the  constant  continuous  current  and  the  in- 
stantaneous alternating  current. 

*  Equation  (16)  is  the  resultant  of  n  magnetomotive  forces  each  of  which  has  the 
maximum  value  —  ampere-turns,  and  having  a  displacement  of  —  degrees  in  both  space 
(direction)  and  time. 

f  The  two-ring  or  single-phase  converter  has  a  very  limited  commercial  application. 

t  See  Chapter  8,  Section  9. 


174  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

Let  the  position  of  the  armature  of  a  two-ring  two-pole  converter 
be  such  that  the  angle  co/  is  zero.  Fig.  140.  Assuming  unity  power 
factor,  the  current  in  the  alternating-current  system  is  maximum 
and  its  direction  of  flow  opposite  to  that  of  the  current  flow  in  the 
continuous-current  circuit.  The  current  flowing  in  the  armature 
windings  is,  therefore,  the  algebraic  sum  of  the  constant  continuous 
current  and  the  maximum  alternating  current. 

When  the  armature  has  advanced  thirty  degrees,  the  alternating 
current  has  decreased  in  value  and  the  continuous  current  in 
coils  a  and  g  has  reversed  in  direction,  the  coils  having  passed  to 
the  opposite  sides  of  the  brushes.  The  current  flowing  in  coils  a 
and  g  is,  therefore,  the  arithmetical  sum  of  the  constant  continuous 
current  and  the  instantaneous  value  of  the  alternating  current,  and 
is  greater  than  in  the  other  coils.  Similarly,  when  the  armature 
has  rotated  through  another  thirty  degrees,  the  direction  of  the 
continuous  current  in  coils  b  and  h  has  reversed,  and  the  current  in 
coils  a,  b,  g  and  h  is  the  arithmetical  sum  of  the  constant  continuous 
current  and  the  instantaneous  alternating  current.  When  the 
armature  has  rotated  through  ninety  degrees,  the  value  of  the  alter- 
nating current  is  zero  and  the  current  flowing  in  any  coil  is  one-half 
that  in  the  continuous-current  (armature)  circuit. 

During  the  next  quarter  revolution  the  alternating  current 
increases  in  the  opposite  direction,  and  coils  e,  f,  I  and  m  carry  the 
larger  instantaneous  current  values. 

From  the  above  it  is  evident  that: 

(a)  The  armature  coils  of  a  two-ring'converter  are  not  uniformly 
heated. 

(b)  The  heating  decreases  as  the  distance  from  the  point  of  con- 
nection to  the  ring  increases. 

The  conditions  shown  to  exist  in  a  two-ring  converter  may  be 
shown  to  exist  in  one  having  a  larger  number  of  rings,  an  increase 
in  the  number  of  rings  making  the  heating  more  nearly  uniform 
and  decreasing  the  total  quantity  of  heat  liberated  in  the  armature 
for  a  given  value  of  load  (continuous)  current. 

Taking  as  the  axis  the  line  AB  in  Fig.  140,  the  instantaneous 
current  in  any  armature  coil  of  an  w-ring  two-pole  converter  is 

2  I  COS  (eo/  =b  a)        I  ,   0, 

I  = >  (10) 

.      7T  2 

wsm- 
n 


CURRENT-RECTIFYING  APPARATUS  175 

and 

.»2  =  -  r?d(<*£)  (19) 

7T  J 

(20) 


when   7  =  the  continuous-current  output, 

n  =  the  number  of  rings  on  the  armature, 

a  =  the  angular  displacement  of  the  coil  from  a  position 

midway  between  the  connections   to   two  adjacent 

rings. 

But  the  heating  of  the  coil  due  to  the  continuous  current  alone  is 

72 
proportional  to  —  .     Therefore, 

4 


l6cosa    ,       \ 


7T  .      7T  "    j  (2l) 


HIT  sin  - 
n 


} 

n        I 


(22) 


when  ^;  =  the  ratio  of  the  average  rate  at  which  heat  is  liberated 
in  the  coil  of  an  w-ring  converter  to  the  rate  at  which  heat  is  liber- 
ated in  the  same  coil  of  the  armature  when  operated  as  a  continuous- 
current  generator  with  the  same  current  output.  Equation  (22) 
is  independent  of  the  number  of  poles  for  which  the  armature  is 
wound,  and  is,  therefore,  applicable  to  multipolar  machines. 

Integrating  equation  (22)  with  respect  to  a  from  a.  =  o  to  <*  =  - 

n 


(24) 


when  k  is  the  ratio  of  the  average  heat  liberated  in  the  armature 
winding  of  an  w-ring  converter  to  the  heat  liberated  in  the  same 


176  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

armature  when  used  as  a  continuous-current  generator,  with  the 
same  current  output. 

20.  Converter  ratings.  —  From  the  above  considerations  it  is 
evident  that  the  current  output  (rating)  of  a  given  armature  when 
used  as  a  synchronous  converter  is  to  the  current  output  (rating) 
when  used  as  a  continuous-current  generator  as  i  is  to  Vk 

(25) 
(26) 


7T 


when  KW  =  the  rating  as  a  continuous-current  generator, 
K'W'  =  the  rating  as  a  synchronous  converter. 

When  the  power  factor  of  the  converter  is  less  than  unity,  the 
alternating  current  is  increased  for  a  given  continuous-current 
output,  the  inequality  of  the  heating  of  the  armature  is~  increased, 
the  heating  of  the  individual  armature  coils  becomes  more  pro- 
nounced, and  the  wattless  component  of  current  tends  to  magnetize 
or  to  demagnetize  the  field  as  the  current  leads  or  lags  behind  the 
electromotive  force. 

CHAPTER  X  —  PROBLEMS 

1.  The  continuous  voltage  of  a  synchronous  converter  is  550.     Find  the 
alternating  voltage  (neglecting  resistance  drop)  between  adjacent  rings  when 
the  converter  is  provided  with:  (a)  v  rings,  (b)  3  rings,  (c)  4  rings,  (d)  6  rings. 

2.  Calculate  the  current  in  the  alternating-current  mains  when  the  output 
of  the  converter  in  Problem  i  is  1000  amperes,  the  efficiency  95  per  cent  and 
the  power  factor  96  per  cent. 

3.  The  power  factor  of  a  2-ring  converter  is  unity  and  the  continuous- 
current  output  250  amperes.    Determine  the  maximum  (instantaneous)  cur- 
rent flowing  in  a  conductor:  (a)  adjacent  to  the  connection  from  the  armature 
winding  to  the  ring,  (b)  mid-way  between  the  connections  to  the  rings. 

4.  Same  as  Problem  3  except  the  power  factor  of  the  alternating-current 
circuit  is  0.866. 

5.  The  continuous  voltage  of  a  4oo-kw.,  3-ring  converter  is  600,  the  re- 
sistance of  its  armature  (measured  between  continuous-current  brushes)   is 
0.027,  and  the  no-load  input  when  operated  as  a  continuous-current  motor 
without  load  (at  rated  speed  and  normal  field  excitation)  is  12.5  kw.     Calculate 
the  efficiencies  for  25,  50,  75,  100  and  125  per  cent  of  rated  output  and  plot  the 
efficiency  curve  (using  per  cent  of  rated  output  as  abscissas). 


CURRENT-RECTIFYING  APPARATUS  177 

6.  The  converter  in  Problem  5  is  provided  with  6  rings.     Determine  its 
rating,  calculate  the  efficiencies,  and  plot  the  efficiency  curve  as  in  Problem  5. 

7.  A  compound-wound  25-cycle,  3-ring  converter  has  an  inductance  con- 
nected in  each  alternating-current  supply  line.     When  the  input  to  the  conver- 
ter is  100  kw.,  the  power  factor  of  the  converter  is  unity  and  the  voltage  between 
continuous-current  brushes  is  600.     When  the  input  to  the  converter  is  10  kw. 
the  power  factor  of  the  converter  is  0.5  and  the  voltage  between  continuous- 
current  brushes  is  550.     Find:  (a)  the  value  of  the  inductance  in  the  alternating- 
current  lines,  (b)  the  voltage  of  the  alternating-current  supply  system. 

8.  The  no-load  continuous  voltage  of  a  split-pole  converter  is  550.     The 
regulating  pole  is  excited  by  means  of  a  series  winding,  and  the  flux  pro- 
duced by  the  regulating  pole  at  full  load  is  30  per  cent  of  the  flux  produced  by 
the  main  winding  which  is  constant.    The  distance  from  the  center  line  of  the 
main  pole  to  the  center  line  of  the  regulating  pole  is  45%  of  the  distance  between 
the  center  lines  of  adjacent  main  poles.     Find  the  full-load  continuous  voltage, 
the  applied  electromotive  force  at  the  rings  remaining  constant. 

9.  A  6-pole,  lap-wound  armature  has  720  conductors.    The  commutator 
has  360  bars.     Determine  the  commutator  bars  which  must  be  connected  to  a 
given  ring  in:  (a)  a  2-ring  converter,  (b)  a  3-ring  converter,  (c)  a  4-ring  con- 
verter, ((/)  a  6-ring  converter. 


CHAPTER  XI 


FIG.  144.     Schematic  Diagram 
of  Transformer. 


THE   TRANSFORMER 

A  TRANSFORMER  consists  of  a  single  magnetic  circuit  linking  with 
two  independent  electric  circuits,  as  shown  schematically  in  Fig. 
144.  Since  iron  offers  a  path  of  small  reluctance  for  the  magnetic 

flux,  the  coils  of  commercial  transformers 
are  always  wound  on  iron  cores. 

i.  Fundamental  physical  action.  — 
When  the  armature  conductors  of  a  con- 
tinuous-current motor  move  across  the 
magnetic  field,  a  counter-electromotive 
force  which  opposes  the  flow  of  current 
is  induced  in  the  conductors,  and  the  armature  rotates  at  a  speed 
which  establishes  equilibrium  in  the  system.  Similarly,  when  an 
alternating  current  flows  in  the  windings 
of  coil  P  (Fig.  144)  the  changing  value 
of  the  flux  induces  *  in  the  coil  an  elec- 
tromotive force  which  opposes  the  flow 
of  current,  and  the  current  is  of  a  value 
which  establishes  equilibrium  in  the  sys- 
tem. The  induced  electromotive  force 
is  equal  to  the  geometrical  difference  of 
the  applied  electromotive  force  and  the 
drop  due  to  the  resistance  of  the  coil. 
The  resistance  drop  in  the  primary 
windings  of  commercial  transformers  FlG*  I45 
operating  without  load  (secondary  cir- 
cuit open)  is  so  small  as  to  be  negligible,  and  the  applied  and  the 
counter-electromotive  forces  are,  therefore,  equal. 
Referring  to  Fig.  144,  let 

0m  =  the  maximum  flux  produced  in  the  iron  core  by  coil  P, 

f  =  the  frequency  of  the  alternating-current  supply  circuit, 
Np  =  the  number  of  series  turns  in  coil  P, 
Nt  =  the  number  of  series  turns  in  coil  5, 

*  See  Chapter  2,  Section  13. 
178 


Transformer  in  Posi- 
tion on  Pole. 


THE  TRANSFORMER  179 

The  flux  in  the  iron  must  decrease  from  its  maximum  <f>m  to  zero 
in  one-fourth  of  a  cycle.  The  average  rate  of  flux  change  is,  there- 
fore, 


d) 

4? 

and  the  average  electromotive  force  induced  in  coil  P  is  the  prod- 
uct of  the  average  rate  of  change  in  the  flux  and  the  number  of 
series  -turns  in  the  coil. 

Eav  =  4<i>mfNP  io-8.  (2) 

Since  the  ratio  of  the  effective  electromotive  force  to  the  average 
electromotive  force  is  i.n, 

Ep  =  4.44<t>mfNP  io-8.  (3) 

Assuming  that  the  flux  set  up  by  coil  P  is  confined  to  the  iron 
core,  the  electromotive  force  induced  in  coil  S  is 

E,  =  444<t>mfN,  io-8.  (4) 

If  the  terminals  of  coil  5  are  connected  through  a  resistance  as 
shown  in  the  figure,  current  flows  in  the  windings  of  the  coil.  The 
current  in  coil  5  tends  to  establish  in  the  iron  core  a  flux  opposed 
to  that  set  up  by  coil  P  *  and  the  counter-electromotive  force  in- 
duced in  coil  P  is  reduced.  The  equilibrium  of  the  system  is  thus 
destroyed,  and  the  current  in  coil  P  increases  until  it  neutralizes 
the  magnetic  effect  of  coil  5. 

As  the  current  in  coil  P  increases,  the  resistance  drop  is  no 
longer  negligible  and  equilibrium  is  reestablished  with  a  counter- 
electromotive  force  less  than  the  applied  electromotive  force. 
Equilibrium  in  a  transformer  is,  then,  maintained  by  an  automatic 
change  in  the  current  flowing  in  the  primary  coil.  The  conditions 
are  analogous  to  those  in  a  shunt  motor,  the  speed  of  which  decreases 
as  the  load  increases,  so  that  the  torque  developed  in  the  motor 
armature  is  automatically  adjusted  to  equal  the  counter  torque 
due  to  the  load. 

2.  Magnetic  leakage.  —  In  Section  i  it  is  assumed  that  all  the 
flux  set  up  by  coil  P  passes  through  coil  S.  Such  is  not  the  case  in 
a  transformer  since  the  flux  is  not  confined  to  the  iron.f  That  part 

*  See  Chapter  2,  Section  14. 
t  See  Chapter  2,  Section  21. 


l8o  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

of  the  flux  which  threads  one  coil  but  does  not  thread  the  other  is 
termed  "  leakage  "  flux,  and  has  the  same  effect  as  a  series  inductance, 
i.e.,  it  reduces  the  electromotive  force  induced  in  coil  S,  and  causes 
the  primary  current  to  lag  behind  the  applied  electromotive  force. 
3.  Electromotive  force  relations.  —  If  the  coils  had  no  resistance 
and  there  were  no  magnetic  leakage,  it  is  evident,  from  equations 
(3)  and  (4),  that  the  ratio  of  the  voltage  applied  at  the  terminals 
of  coil  P  to  the  voltage  delivered  at  the  terminals  of  coil  S  would 
be  that  of  the  number  of  turns  in  the  coils. 

,  , 
Ea  ~8 


This  ratio  is  found  to  hold  very  closely  when  the  secondary  circuit 
is  unloaded.  Because  of  resistance  and  magnetic  leakage,  the 
secondary  terminal  voltage  decreases  as  the  load  increases,  the 
applied  voltage  remaining  constant. 

4.  Current  relations.  —  Neglecting  the  magnetizing  current  and 
the  losses,  which  are  small,  the  input  to  a  transformer  is  equal  to  its 
output,  and  the  product  of  the  electromotive  force  and  the  current 
in  the  primary  circuit,  is  equal  to  the  product  of  the  electromotive 
force  and  the  current  in  the  secondary  circuit. 

EPTP  =  EJ8.  (7) 

Therefore 

y*  =  f.  (8) 

1,        tLv 

N.  . 


i.e.,  the  ratio  of  the  primary  and  the  secondary  currents  is  equal 
to  the  inverse  ratio  of  the  number  of  series  turns  in  the  coils. 

5.  Vector  diagram.  —  The  electromotive  forces  and  currents  in  a 
transformer  may  be  represented  graphically  by  means  of  a  vector 
diagram.  Fig.  i46a  is  the  diagram  of  an  unloaded  transformer. 
The  applied  electromotive  force  is  represented  by  OA,  the  secondary 
electromotive  force  by  OB  and  the  flux,  90  degrees  behind  OA 
and  90  degrees  ahead  of  OB,  by  OC.  Because  of  hysteresis  in 
the  iron  core,  the  no-load  current  is  not  harmonic,*  and  cannot, 
*  See  Appendix  B,  Section  8. 


THE  TRANSFORMER 


181 


therefore,  be  represented  by  a  rotating  vector.  It  is  represented  in 
transformer  diagrams  by  what  is  termed  the  "  equivalent  harmonic 
current,"  i.e.,  a  harmonic  current  having  the  same  frequency  and 
effective  value  as  the  actual  current.  The  error  due  to  this  assump- 


- 


FIG.  i46a.    Vector  Diagram  of  Unloaded 
Transformer. 


FIG.  1466.    Vector  Diagram  of  Loaded 
Transformer. 


tion  is  usually  negligible.  If  the  resistance  of  coil  P  were  zero  and 
there  were  no  iron  (hysteresis  and  eddy  current)  losses,  the  no-load 
current  would  be  in  phase  with  the  flux.  Because  of  these  losses, 
which  are  practically  all  iron  losses,  there  is  a  power  component  of 
the  no-load  current  in  phase  with  the  applied  electromotive  force 
OA,  and  the  vector  of  the  no-load  current  is  less  than  90  degrees 
behind  OA. 

That  the  electromotive  force  induced  in  the  coils  of  a  transformer 
is  90  degrees  out  of  phase  with  the  flux  set  up  by  the  primary  wind- 
ing is  evident  from  the  following  consideration: 

Let  the  flux  in  the  iron  core  (Fig.  144)  vary  harmonically.  The 
electromotive  force  induced  in  coils  P  and  5  is  proportional  to  the 
rate  of  change  in  the  flux  threading  the  coils.*  But  the  rate  of 
flux  variation  is  maximum  when  the  flux  is  passing  through  its  zero 
value,  f 

When  the  secondary  circuit  is  loaded,  the  transformer  quantities 
are  represented  in  Fig.  i46b.  Let 

Ep  =  the  applied  electromotive  force, 

Ea  =  the  secondary  (terminal)  electromotive  force, 

IP  =  the  current  in  the  primary  circuit, 

I8  =  the  current  in  the  secondary  circuit, 

*  See  Chapter  2,  Section  13. 
t  See  Appendix  A,  Section  7. 


182  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

Rp  =  the  resistance  of  the  primary  circuit, 
Rs  =  the  resistance  of  the  secondary  circuit, 
Xp  =  the  reactance  of  the  primary  circuit, 
Xs  =  the  reactance  of  the  secondary  circuit, 
cos  <f>  =  the  power  factor  of  the  primary  circuit. 

Draw 

OA  =  Ep, 
OD  =  Ip, 
(f>  =  AOD. 

The  applied  voltage  Ep  is  equal  to  the  geometric  sum  of  the  resis- 
tance drop  Rplp,  the  reactance  drop  XPIP  and  the  counter-electro- 
motive force. 
Draw 

AG  =  Xplp  perpendicular  to  the  current  vector  OD, 
GE  =  RpIP  parallel  to  the  current  vector. 

Then  OE  is  the  vector  of  the  counter-electromotive  force  induced 
in  the  primary  winding.  . 

The  load  component  of  the  primary  current  OD'  is  the  geometric 
difference  of  OD  and  the  no-load  current. 

The  secondary  current  Is  is  in  phase  opposition  to  the  primary 
load  current  OD'  . 

The  electromotive  force  induced  in  the  secondary  winding  is  in 
phase  opposition  to  OE. 

Draw 

OK  =  Is, 


OF  =  OE-s, 

NP 

FH  =  XJS  perpendicular  to  OK, 
BH  =  RJ8  parallel  to  OK. 

OB  is  the  secondary  terminal  voltage  Es. 

6.  Types  and  construction  of  transformers.  —  Commercial  con- 
stant-potential transformers  differ  in  the  mechanical  arrangement 
of  the  iron  core  and  the  windings,  and  are  of  two  general  types: 
(a)  core,  (6)  shell. 

(a)  Core  type.  —  In  the  core  type  transformer  the  windings  are 
placed  around  the  legs  of  the  core  and  cover  a  large  part  of  the 
iron.  Fig.  i4ya. 


I 


THE  TRANSFORMER 


(b)  Shell  type.  —  In  the  shell  type  transformer  the  windings 
are  placed  around  the  middle  leg  of  the  core  as  shown  in  Fig.  1470. 
Iron,  evidently,  encloses  the  greater  part  of  the  coils. 

There  is  no  marked  superiority  of  one  type  over  the  other, 
both  types  being  largely  used.  Economical  considerations  usually 


FIG.  i47a.   Core  Type  Transformer  (Case     FIG.  i47b.    Shell  Type  Transformer  (Case 
Removed).     Maloney  Electric  Co.  Removed).    General  Electric  Co. 


result  in  the  selection  .of  the  shell  type  when  the  voltages  are  low 
and  the  currents  relatively  large,  and  of  the  core  type  when  the 
voltages  are  high  and  the  currents  relatively  small. 

Transformer  cores  are  built  up  of  thin  stampings  (laminations) 
to  reduce  eddy-current  losses.  After  the  core  and  the  windings 
are  assembled  they  are  placed  in  an  iron  case  for  mechanical 
protection. 

7.  Transformer  losses.*  —  The  losses  in  a  transformer  are:  (a) 
iron  losses,  (b)  copper  losses. 

(a)  Iron  losses.  —  The  iron  losses  of  a  transformer  are  those 
due  to:  (i)  hysteresis,  (2)  eddy-currents. 

(i)  Hysteresis  loss.  —  The  hysteresis  loss  in  an  iron  core  f  is 
equal  to 

P,  =  i,F/B.M,  (10) 


*  See  Standardization  Rules  of  the  American  Institute  of  Electrical  Engineers, 
t  See  Appendix  B,  Section  7. 


184  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

when  rj  =  the  magnetic  (hysteretic)  constant  of  the  iron  used  in 

the  core, 

V  =  the  volume  of  the  iron, 

/  =  the  frequency  of  the  applied  electromotive  force, 
Bm  =  the  maximum  flux  density  in  the  iron. 

Since,  in  any  given  transformer,  77,  V  and  /  are  constant,  and  Bm 
is  proportional  to  the  applied  electromotive  force,  equation  (10) 
may  be  written 

P»  =  *»£".  (n) 

(2)  Eddy-current  losses.  —  The  eddy-current  loss  in  an  iron  core  * 

(12) 


when  b  =  a  constant  proportional  to  the  reluctivity  of  the  iron  used 

in  the  core, 

V  =  the  volume  of  the  iron, 

/  =  the  frequency  of  the  applied  electromotive  force, 
/  =  the  thickness  of  the  laminations  of  which  the  core  is  built. 
Bm  =  the  maximum  flux  density  in  the  iron. 

Equation  (12)  may  be  written 

p.  =  ke&.  (i3) 

Determination  of  total  iron  loss.  —  The  total  iron  loss  in  a  trans- 
former is  determined,  for  any  applied  voltage,  by  connecting  the 

-  -      transformer  as  in  Fig.   148,  and 
impressing  the  required   voltage 
across  its  terminals.     The  watt- 
_  .      meter    indicates    the    total    iron 

FIG.  148.     Connections  for  Transformer     Joss  pjus    the   copper   loss   due   to 

Iron  Losses.  ^  no_load  current.  The  no-load 

copper  loss  is  usually  so  small  as  to  be  negligible,  but  the  correction 
is  easily  made  after  the  resistance  of  the  primary  winding  has  been 
determined. 

Separation  of  hysteresis  and  eddy-current  losses.  —  The  iron  loss 
of  a  transformer  is  separated  into  its  components  by  the  following 
graphical  method:  From  equations  (10)  and  (12) 


*  See  Chapter  6,  Section  i. 


Supply 


THE  TRANSFORMER 


18* 


when  B  is  constant,  i.e.,  when  the  ratio  —  is  constant.     Dividing 
equation  (14)  by/ 

-;  =  kh'  +  kSf,  (15) 


which  is  the  equation  of  a  straight  line,  and  may  be  plotted  as  ab, 
Fig.  149.     The  ordinate  oc  is  the  value  of  the  constant  kh'. 


kef 


0  10  20          30          40  50          60         70 

FREQUENCY  (f ) 

FIG.  149.    Separation  of  Hysteresis  and  Eddy-current  Losses. 


The  power  lost  in  hysteresis  at  any  frequency  is  obtained  by  multi- 
plying the  ordinate  oc  =  khf  by  the  frequency;  and  the  power  lost 
in  eddy  currents  is  obtained  by  multiplying  the  ordinate  ke'f,  corre- 
sponding to  the  required  frequency,  by  the  frequency.* 

(b)  Copper  losses.  —  The  copper  losses  in  a  transformer  are  those 
due  to  the  resistance  of  the  windings,  and  are  equal  to  the  product 
of  the  resistance  and  the  square  of  the  current. 


PC  =  RPIP2  +  RJ*, 


(16) 


*  Compare  with  the  separation  of  hysteresis  and  eddy-current  1 
current  machinery  as  outlined  in  Chapter  6,  Section  4. 


in  continuous- 


l86  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

when   Pc  =  the  total  copper  loss  in  the  transformer, 
Rp  =  the  resistance  of  the  primary  winding, 
R8  =  the  resistance  of  the  secondary  winding, 
IP  =  the  primary  current, 
I8  =  the  secondary  current. 

The  resistance  of  a  transformer  winding  is  determined  by  con- 
necting it,  in  series  with  a  suitable  rheostat,  to  a  source  of  contin- 
uous current,  and  measuring  the  current  in  the  circuit  and  the 
voltage  between  the  terminals  of  the  winding.  The  resistance  of 
the  winding  is,  from  Ohm's  Law, 


Direct  measurement  of  the  copper  losses  may  be  made  by  con- 
necting a  transformer  as  indicated  in  Fig.  150,  where  the  low- 
voltage  winding  is  short-circuited  and  the  high-voltage  winding 

connected  to  a  source  of  alternat- 
ing current,  and  the  voltage  be- 
tween the  terminals  of  the  winding 
regulated  to  give  any  required 

FIG.  150.     Connections  for  Transformer    Current  in  the  windings.    The  volt- 
Copper  Losses.  age  required  to  produce  full-load 

current  in  the  windings  of  a  short-circuited  transformer  is  usually 
only  a  few  per  cent  of  the  rated  voltage  of  the  coil.  The  indication 
of  the  wattmeter  is  the  sum  of  the  total  copper  loss  in  the  trans- 
former and  the  iron  loss  in  the  core.  Since  the  iron  loss  in  a 
transformer  operating  at  rated  voltage,  is  seldom  more  than  3 
per  cent  of  its  rated  output,  the  iron  loss  at  the  reduced  voltage 
required  for  the  short-circuit  test  is  negligible,  i.e.,  the  indication  of 
the  wattmeter  is  the  total  copper  loss  in  the  windings. 

8.  Equivalent  resistance.  —  The  equivalent  resistance  of  a 
transformer  is  the  resistance  which,  when  multiplied  by  the  square 
of  the  current  in  the  circuit,  gives  the  total  copper  loss  of  the  trans- 
former. It  will,  evidently,  have  different  values  as  the  current 
used  is  that  of  the  primary  or  of  the  secondary  circuit.  Let 

Re  =  the  equivalent  resistance  of  a  transformer, 
Rp  =  the  resistance  of  the  primary  winding, 
Ra  =  the  resistance  of  the  secondary  winding, 
IP  =  the  primary  current, 


THE   TRANSFORMER  187 

Is  =  the  secondary  current, 
Pp  =  the  copper  losses  in  the  primary  winding, 
Ps  =  the  copper  losses  in  the  secondary  winding, 
Pc  =  the  total  copper  losses  in  the  transformer. 

Then 

Pc  =  Pp  +  P*  (18) 


+  RJ*.  (19) 

From  equation  (9) 

''-'•£  (20) 

and 

/.-/,£  («) 

Therefore, 


From  equation  (22) 

'  (24) 


when  referred  to  the  primary  circuit.     From  equation  (23) 

-Y  +  R.,  (25) 


when  referred  to  the  secondary  circuit. 

9.  Equivalent  reactance.  —  The  equivalent  reactance  of  a  trans- 
former is  the  ratio  of  the  voltage  drop  due  to  the  inductance  of 
the  windings  and  to  leakage  flux,  and  the  current  flowing  in  the 
circuit.  It  is  determined  from  the  short-circuit  test  and  the 
equivalent  resistance  of  the  windings. 


when  Xe  =  the  equivalent  reactance  of  the  transformer,  referred 
to  the  primary  or  to  the  secondary  circuit  as  the 
values  of  E,  I  and  R  are  primary  or  secondary 
values, 


l88  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

E  =  the  value  of  the  primary,  or  the  secondary,  electro- 
motive force  required  to  cause  current  /  to  flow  in 
the  primary  circuit,  or  in  the  short-circuited  windings. 
/  =  the  current  flowing  in  the  primary  circuit,  or  in  the 
short-circuited  windings. 

Re  =  the  equivalent  primary,  or  secondary,  resistance  of 
the  windings. 

10.  Cooling.  —  The  cases  of  small  transformers  are  filled  with 
oil  which  helps  dissipate  the  heat  liberated  in  the  windings  and 
in  the  core.     In  large  transformers  additional  means  of  cooling 
are  required,  the  heat  being  dissipated  by  means  of  an  air  blast,  or 
by  means  of  a  system  of  pipes  through  which  cold  water  is  forced, 
and  around  which  the  oil  circulates. 

It  is  cheaper  to  cool  large  transformers  by  mechanical  means 
than  to  build  them  of  such  "shape  and  dimensions  that  they  are 
self-cooling. 

11.  Efficiency.  —  The  efficiency  of  a  properly  designed  trans- 
former is  high,  and  varies  from  95  per  cent  in  small  transformers  to 
above  99  per  cent  in  those  of  large  size.     If  the  applied  electromo- 
tive force  is  constant,  the  iron  loss  is  approximately  constant,  the 
copper  loss  is  proportional  to  the  square  of  the  load  and  the  effi- 
ciency may  be  calculated  from  the  following  equation: 

Per  cent  efficiency  =  — - ~~JT^J* '  (2?) 

E8IS  cos  <j>  +  P0  +  Rel8 

when  Ea  =  the  rated  secondary  voltage, 

I8  =  the  current  output  at  which  it  is  desired  to  calculate  the 

efficiency, 

P0  =  the  no-load  input  (=  iron  losses), 
Re  =  the  equivalent  secondary  resistance  of  the  windings, 
cos  <£  =  the  power  factor  of  the  load  circuit. 

Transformer  efficiencies  are  usually  calculated  for  unity  power 
factor,  under  which  condition  cos  0  in  equation  (27)  is  equal 
to  i. 

Example.  —  Find  the  full-load  efficiency  of  a  2200:  2  20- volt,  15- 
kv-a.  transformer,  the  no-load  input  to  which  is  200  watts,  RP  —  2 
ohms,  Rs  =  0.02  ohm. 


THE  TRANSFORMER  189 

Ea  —  220  volts, 
15,000  =  6g  x         eres 

220 

P0  =  200  watts, 
Re=  0.02  -f-  2  X  -  -  =  0.04  ohm, 

100 

RJ*  =  0.04  X  (68.i)2  =  185.5  watts> 

COS  (j>  =  I . 

~  .  15,000  X  100 

Per  cent  efficiency  = 


15,000  -f-  200  +  185.5 

I5*000 
~  15,385-5 
=  97-5- 

The  "all-day  efficiency"  of  a  transformer  is  the  ratio  of  the  total 
output  during  twenty-four  hours  to  the  total  input  during  the 
same  period. 

~,  .  output  during  24  hours  /  ox 

All-day  efficiency  =  -  (28) 

input  during  24  hours 

output  during  24  hours 

output  during  24  hours  +  losses 

Example.  —  Calculate  the  all-day  efficiency  of  the  above  trans- 
former, when  operated  at  full  load  for  8  hours  each  day. 

rr     •  15,000   X8    X    100 

Per  cent  efficiency  =  - 

15,000  X  8  -f  200  X  24  +  185.5  X  8 

_  120,000 
126,248 

=  95- 

12.  Regulation.  —  The  regulation  of  a  transformer  is  the  ratio 
of  the  increase  e  in  the  secondary  terminal  voltage,  between  rated 
load  and  no  load,  and  the  secondary  voltage  E8  at  rated  load,  the 
primary  voltage  remaining  constant. 

Per  cent  regulation  —  ^-^  •  t-=  (30) 

The  secondary  terminal  voltage  decreases  as  the  load  increases 
because  of:  (a)  the  resistance  of  the  windings,  (b)  the  reactance  of 


190  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

the  windings.  Since  the  vector  sum  of  the  secondary  terminal 
voltage,  the  resistance  drop  and  the  reactance  drop  is  constant,  the 
regulation  of  a  transformer  becomes  larger  (poorer)  as  the  power 
factor  of  an  inductive  load  circuit  decreases. 

The  resistance  and  the  reactance  of  a  transformer  are  determined 
as  explained  in  Sections  8  and  9,  and  the  regulation  calculated  by 
means  of  the  following  equation: 
Per  cent  regulation 


V  \£LSCQS  r    , ,     — , T    .  --„-„,         — a  ^, 

~jg—  -XlOO,  ^i; 

when  Es  =  the  full-load  (rated)  secondary  voltage, 
I8=  the  full-load  (rated)  secondary  current, 
Re  =  the  equivalent  secondary  resistance, 
Xe=  the  equivalent  secondary  reactance, 
cos  0  =  the  power  of  the  load  circuit. 

.  Example.  —  Determine  the  regulation  of  the  transformer  in 
Section  n,  if  no  volts  must  be  applied  to  the  2 200- volt  winding 
to  cause  rated  current  to  flow  in  the  short-circuited  secondary 
winding. 

E5  =  220  volts, 

I8=  68.1  amperes, 

Re  =  0.04  ohms, 

//H\2 

Xe=  Y  (— j  +  (o.o4)2  =  0.16  ohms, 

cos  0  =  i, 

RJ8  =  0.04  X  68.1  =  2.7  volts, 
XeI8  =  0.16  X  68.1  =  10.9  ohms. 

-n  i    -L-  V(22O  +  2.7)2  +  (lO.o)2  —  22O 

Per  cent  regulation  =  — *—  -  X  100 

220 


22O 
=   1.4. 


13.   The  auto-transformer.  —  When  a  single  continuous  winding 
forms   both   the  primary  and   the   secondary  of   a   transformer 


THE  TRANSFORMER 


the  arrangement  is  termed  an  auto- transformer.  Fig.  151.  The 
auto-transformer,  like  the  independent  coil  transformer,  is  rever- 
sible and  may  be  used  to  increase  or  to  decrease  the  applied  voltage. 

The  ordinary  (two-winding)  transformer 
may  be  used  as  an  auto-transformer  by 
interconnecting  the  windings  as  indicated 
in  Fig.  152.  Let 

£1  =  220  volts, 


Nt 


=  10, 


10  amperes. 


FIG.  151.     Conventional  Dia- 
gram of  Auto-Transformer. 


When  the  coils  are  interconnected  as  indicated  in  Fig.  i52a,  the 
voltage  of  the  load  circuit  is  the  arithmetical  sum  of  the  applied 
voltage  and  the  electromotive  force  induced  in  the  secondary 

winding;  when  connected  as  indi- 
cated in  Fig.  i52b,  the  load  voltage 
is  the  arithmetical  difference  of  the 
applied  voltage  and  the  electromo- 
tive force  induced  in  the  secondary 
winding.  The  above  statements  are 
evident  from  the  fact  that  the  wind- 
ings  may  be  interconnected  so  that 
the  applied  and  the  secondary  (in- 
duced)  electromotive  forces  are  either 
in  phase  or  in  phase  opposition. 
The  ratio  of  the  currents  in  the 

windings   of   an  auto-transformer  is 
FIG.  152.    The  Auto-Transformer. 

the  same  as  if  the  transformer  were 

operating  in  the  usual  manner,  i.e.,  the  currents  in  the  windings 
are  inversely  proportional  to  the  number  of  turns  in  the  windings, 
while  the  current  in  the  supply  mains  is  the  algebraic  sum  of  the 
currents  in  the  two  windings.  Assuming  the  load  circuit  to  be 
non-inductive,  the  power  delivered  by  the  transformer  when  con- 
nected as  indicated  in  Fig.  i52a,  is 

P  =  242  X  10 
=  2420  watts. 

The    same    power,    neglecting    the    losses,    is    supplied    to    the 


19  2  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

transformer,  but  the  voltage  of  the  primary  circuit  is  only  220. 
Therefore, 


22O 

=  11  amperes, 

and  the  current  flowing  in  the  primary  winding  is  i  ampere. 

Similarly,  for  the  connection  indicated  in  Fig.  1520,  the  power 
delivered  is 

P  =  198  X  10 
=  1980  watts 
ii 


220 

=  9  amperes, 

and  the  current  flowing  in  the  primary  winding  is  i  ampere. 

The  copper  losses  in  an  auto-transformer  are  less  than  in  a  two- 
circuit  transformer  having  the  same  output  and  the  same  primary 
and  secondary  voltages,  but  the  interconnection  of  the  windings 
makes  its  use  inadvisable  except  where  the  difference  between  the 
supply  voltage  and  that  of  the  load  circuit  is  small.  One  of  its 
principal  uses  is  to  reduce  the  voltage  applied  to  an  alternating- 
current  motor  during  the  period  of  acceleration. 

14.  The  constant-current  transformer.  —  A  transformer  having 
a  counterbalanced  movable  coil  delivers  approximately  constant 
current  to  a  load  circuit  of  variable  impedance,  when  the  voltage 
between  the  terminals  of  the  primary  winding  is  constant. 

When  an  alternating  current  flows  in  coil  P  the  magnetic  flux 
set  up  divides  between  the  iron  and  the  air  as  indicated  in  Fig.  153. 
That  part  of  the  flux  which  threads  coil  S  induces  in  it  an  electro- 
motive force  which  causes  a  current  to  flow  in  the  closed  secondary 
circuit.  The  flux  which  passes  through  the  air  forms  a  magnetic 
field,  the  direction  of  which  is  at  right  angles  to  the  current-carry- 
ing conductors  of  coil  5.  Since  the  current-carrying  conductors 
of  coil  S  lie  in  a  magnetic  field,  the  coil  tends  to  move  and  the 
direction  of  its  motion  is  upward.  The  magnetic  circuit  is  designed 
for  large  magnetic  leakage,  so  that  the  number  of  magnetic  lines 
threading  coil  5  is  materially  reduced  as  the  coil  moves  upward. 
Both  the  induced  electromotive  force  and  the  current  flowing  in 
the  coil  are  reduced,  and  the  force  producing,  or  tending  to  produce, 
motion  becomes  less. 


THE  TRANSFORMER 


193 


FIG.  153.    The  Constant-current 
Transformer. 


By  partially  counterbalancing  coil  S  as  shown  in  Fig.  153,  the 
secondary  comes  to  rest  in  that  position  which  establishes  equilib- 
rium in  the  system,  i.e.,  coil  S  moves  upward  until  the  reaction  of 
the  magnetic  field  and  the  current- 
carrying  conductors  is  just  suffi- 
cient to  neutralize  the  effect  of  the 
unbalanced  part  of  the  coil.  If 
the  impedance  of  the  load  circuit 
is  increased,  the  current  in  coil  5 
decreases,  the  upward  force  acting 
on  the  coil  decreases,  and  the  coil 
moves  downward  until  equilibrium 
is  restored,  i.e.,  until  the  flux  thread- 
ing the  secondary  coil  increases 
sufficiently  to  restore  the  original 
value  of  current  flowing  in  the 
secondary  circuit.  If  the  impedance  of  the  load  circuit  is  reduced, 
the  current  in  the  secondary  circuit  increases,  increasing  the  up- 
ward force  on  the  coil,  and  the  coil  rises  until  equilibrium  is  re- 
established. 

As  stated  in  Section  2,  leakage  flux  has  the  same  effect  as  a  series 
inductance,  i.e.,  it  causes  the  current  to  lag  behind  the  applied 
electromotive  force.  The  power  factor  of  a  constant-current 
transformer  is,  therefore,  low  when  the  magnetic  leakage  is 
large,  i.e.,  when  the  transformer  is  operating  at  partial  loads.  In 
fact,  the  primary  current  of  a  constant-current  transformer  is 
approximately  constant,  and  the  change  in  load  is  accomplished  by 
a  change  in  the  power  factor  of  the  primary  circuit.  It  is,  there- 
fore, highly  undesirable  that  these  transformers  operate  at  loads 
much  below  their  rated  output. 

To  make  the  operation  of  constant-current  transformers  at 
partial  loads  commercially  feasible,  taps  are  often  provided  on  the 
secondary  coil,  by  means  of  which  the  number  of  current-carrying 
conductors  in  the  coil  may  be  changed.  The  relative  position  of 
the  coil  for  a  given  load  and,  consequently,  the  leakage  flux,  may 
thus  be  changed,  and  the  power  factor  of  the  system  maintained  at 
approximately  full-load  value.* 

*  The  average  power  factor  of  constant-current  arc  lamp  circuits  is  about  70  per 
cent  when  the  transformers  operate  at  full  load. 


194 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


The  constant-current  transformer  is  largely  used  to  supply  current 
to  arc  and  other  lamps  connected  in  series.     In  large  transformers 

the  coils  may  be  divided,  and 
the  secondary  (movable)  coils 
balanced  against  each  other. 
With  divided  coils  the  size  of 
the  auxiliary  weight  W  is  re- 
duced to  a  minimum,  that 
necessary  to  give  the  required 
range  of  current. 

15.  Polyphase  transform- 
ers.— Polyphase  transformers 
are  constructed  by  interlink- 
ing the  magnetic  circuits  of 
two  or  more  transformers 
which  are  to  be  connected  to 
a  polyphase  system.  If  the 
flux  set  up  by  coil  A  is  in 

FIG.  154.    Constant-current  Transformer  (Case    quadrature  with  that   set  up 
Removed).     General  Electric  Co. 

by  coil  B,  the  weight  of  the 

iron  cores  is  approximately  15  per  cent  less  than  the  weight  of 
the  cores  of  two  equivalent  single-phase  transformers,  the  maximum 
flux  density  being  the  same  in  each  construction.  Polyphase  trans- 
formers are  extensively  used  in  Europe  but  have  not  attained 
great  popularity  in  America. 


CHAPTER  XI  —  PROBLEMS 

1.  The  iron  losses  (measured  at  no-load)  in  a  5-kv-a.  22oo:22o-volt  trans- 
former are  75  watts.     The  resistance  of  the  primary  winding  is  7.5  ohms;  that 
of  the  secondary  winding  is  0.07  ohm.     Calculate  the  full-load  efficiency:  (a) 
when  the  power  factor  of  the  load  circuit  is  unity,  (&)  when  the  power  factor  of 
the  load  circuit  is  0.85. 

2.  The  transformer  in  Problem  i  is  operated  during  24  hours  as  follows: 
4  hours  at  full  load  and  unity  power  factor, 

6  hours  at  half  load  and  90%  power  factor, 
14  hours  without  load. 

Calculate:    (a)  the  all-day  efficiency,  (ft)  the  yearly  loss  charge  against  the 
transformer  if  energy  is  produced  for  i  cent  per  kw.-hour. 

3.  A  20:  i  -transformer,  the  regulation  of  which  is  2  per  cent,  has  a  terminal 
voltage  of  115  at  full  load.    Determine  the  primary  voltage  at  full  load. 


THE  TRANSFORMER  195 

4.  The  hysteresis  loss  in  a  transformer  is  125  watts  when  operated  on  do- 
cycle  mains.     Calculate  the  hysteresis  loss  when  the  transformer  is  operated 
on  a  25-cycle  circuit,  the  voltage  of  which  is  equal  to  that  of  the  oo-cycle  circuit. 

Note.  —  From  equation  (3) 

— f 

5.  The  eddy-current  loss  in  a  transformer  is  100  watts  when  operated  on  60- 
cycle  mains.     Calculate  the  eddy-current  loss  when  the  transformer  is  operated 
on  a  25-cycle  circuit,  the  voltage  of  which  is  equal  to  that  of  the  oo-cycle  circuit. 

6.  20  per  cent  of  the  iron  loss  in  a  6o-cycle  transformer,  when  operated  at 
rated  frequency  and  voltage,  is  due  to  eddy  currents.     Calculate  the  per- 
centage increase  or  decrease  in  the  total  iron  loss  when  the  transformer  is 
operated  at  25  cycles  and  rated  voltage. 

7.  The  hysteresis  loss  in  a  oo-cycle,  2300- volt  transformer  is  500  watts. 
Calculate  the  hysteresis  loss  when  operated  on  a  6o-cycle,  ii5o-volt  circuit. 

8.  The  eddy-current  loss  in  a  6o-cycle,  23oo-volt  transformer  is  400  watts. 
Calculate  the  eddy-current  loss  when  operated  on  a  6o-cycle,  1150- volt  circuit. 

9.  Two  transformers  have  the  same  volume  of  iron  of  the  same  quality, 
but  are  so  wound  that  the  ratio  of  the  flux  densities  is  2 13.     Calculate  the  ratios 
of  the  iron  losses. 

10.  Calculate  the  equivalent  resistance  of  the  transformer  in  Problem  i :  (a) 
referred  to  the  primary  circuit,  (6)  referred  to  the  secondary  circuit. 

11.  Compare  the  iron  losses  in  three  similar  transformers  star-connected  to 
a  balanced  3 -phase  system,  with  the  iron  losses  in  the  same  transformers  delta- 
connected  to  the  same  system. 

12.  Compare  the  copper  losses  in  three  similar  transformers  connected  as  in 
Problem  n,  the  line  current  in  the  star-connected  system  being  the  same  as  the 
line  current  in  the  delta-connected  system. 

13.  An  auto-transformer  has  a  ratio  of  2:3.     Determine  the  relative  size 
of  the  conductors  to  be  used  in  the  windings,  the  current  densities  to  be  equal. 

14.  The  iron  losses  in  a  i5-kv-a.  22oo:22o-volt  transformer  are  135  watts 
(measured  at  no  load).     Resistance  of  primary  winding  =  2  ohms;   resistance 
of  secondary  winding  =  0.018  ohm.     Find:  (a)  the  copper  losses  at  full  load, 
(6)  the  efficiency  at  full  (rated)  load  and  90  per  cent  power  factor,  (c)  the  copper 
losses  when  the  efficiency  is  maximum. 

15.  The  maximum  efficiency  of  a  i2.5-kv-a.  22oo:22o-volt  transformer  is 
97.5  per  cent,  and  occurs  when  the  output  is  80  per  cent  of  the  rated  value. 
Find:  (a)  the  copper  losses,  (b~)  the  iron  losses,  (c)  the  equivalent  resistance  re- 
ferred to  the  2 200- volt  circuit,  (d)  the  equivalent  resistance  referred  to  the  220- 
volt  circuit. 

16.  A  22oo:22o-volt  transformer  is  rated  at  150  kv-a.  and  has  a  no-load 
input  of  1.2  kw.     Primary  resistance  =  0.16  ohm;    secondary  resistance  — 
0.0018  ohm.     Find:  (a)  the  total  copper  loss  at  full  load,  (b)  the  efficiency  at 
full  load,  (c)  the  output  at  maximum  efficiency. 


CHAPTER  XII 
TRANSFORMER   CONNECTIONS 

1.  Protection.  —  In  connecting  transformers  to  a  supply  circuit 
they  should  be  protected  from  excessive  currents  by  means  of 
fuses  of   proper  size  in  the  primary  circuits.     A  fuse  is  a  short 
piece  of  soft  wire  which  melts  and  opens  the  circuit  whenever  an 
excessive  current  flows  in  the  circuit.     Fuse  wire  is  rated  in  amperes 
carrying   capacity,  the   capacity  varying,   approximately,  as  the 
cross-sectional  area. 

2.  For  single-phase  circuits.  —  Connections  for  single-phase  cir- 
cuits are  shown  in  Fig.  155,  where  P  is  the  primary  winding  and 


FIG.  i55a.    Primary  and  Secondary  Coils     FIG.  I55b.     Secondary  Coils  in  Parallel, 
in  Parallel. 


FIG.  I55C.     Primary  Coils  in  Parallel.      FIG.  issd.     Primary  and  Secondary  Coils 

in  Series. 
Single-phase  Transformer  Connections. 

5  is  the  secondary  winding.  In  Figs.  i55a  and  i55b,  the  halves 
of  the  secondary  winding  are  connected  in  parallel,  and  the  volt- 
age of  the  load  circuit  is  that  of  each  coil.  In  making  a  parallel 
connection  care  must  be  taken  that  similar  terminals  of  the  two 
coils  are  connected  together,  or  a  short  circuit  may  be  formed.  In 
Figs.  i55c  and  i55d  the  secondary  coils  are  connected  in  series  and 
the  voltage  of  the  load  circuit  is  twice  that  of  Figs.  i55a  and  i55b, 
the  primary  voltage  being  constant.  It  is  impossible,  in  the  series 

196 


TRANSFORMER   CONNECTIONS  197 

connection,  to  form  a  short  circuit  but  if  the  coils  are  improperly 
connected  their  voltages  oppose  and  neutralize  each  other,  and 
the  voltage  of  the  load  circuit  is  zero. 

Since,  for  a  given  load,  the  product  of  the  current  and  the  elec- 
tromotive force  is  constant,  it  is  evident  that  the  secondary  current 
in  the  parallel  connections  (Figs.  i$5a  and  1550)  is  twice  that  in 
the  series  connections  (Figs.  1550  and  i55d). 

A  three-wire  system  is  formed  by  connecting  a  third  wire  to  the 
junction  of  the  series-connected  coils,  as  indicated  in  Figs.  i55c  and 
i55d.  The  voltage  between  A  and  B  and  that  between  B  and  C  is 
one-half  that  between  A  and  C.  This  connection  offers  two  avail- 
able voltages  and  is  much  used,  the  load  being  connected  between 
any  two  of  the  three  wires.  If  the  load  connected  between  A  and 
B  is  not  equal  to  that  connected  between  B  and  C,  the  halves  of 
the  transformer  are  unequally  loaded.  With  a  properly-distributed 
load,  the  current  in  line  B  is  materially  less  than  that  in  either 
A  or  C.  For  this  reason  B  is  usually  made  smaller  than  A 
and  C. 

Two  or  more  transformers  may  be  connected  so  that  they 
supply  the  same  load  circuit.  In  making  such  connections  the 
same  conditions  must  exist  as  for  the  parallel  operation  of  alterna- 
tors. Since  the  primary  coils  are  connected  to  the  same  supply 
circuit,  the  frequencies  of  the  secondary  circuits  are  the  same, 
but  the  voltages  may  be  in  phase  or  in  phase  opposition.  If  the 
secondaries  are  connected  so  that  their  electromotive  forces  are 
not  in  phase  opposition,  a  short  circuit  is  formed  and  an  excessive 
current  flows  in  the  windings.  This  large  current  heats  the  trans- 
formers to  such  a  degree  as  to  seriously  damage  or  destroy  the 
insulation,  unless  the  protective  devices  work  promptly. 

Transformers  having  different  characteristics  should  not  be 
operated  in  parallel  as  they  will  not  divide  the  load  properly. 
For  example,  if  two  transformers  of  the  same  rating  and  no  load 
voltage,  but  having  regulations  of  2  and  5  per  cent,  are  operated  in 
parallel,  the  one  having  the  better  (smaller)  regulation  will  carry 
the  greater  part  of  the  load.  When  the  total  load  is  equal  to  their 
combined  ratings  both  transformers  operate  at  a  reduced  efficiency, 
one  being  overloaded  and  the  other  underloaded.  The  wave  shapes, 
or  the  ratios  of  transformation  of  different  transformers,  may  be 
such  as  to  make  their  parallel  operation  impossible. 


198  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

3.  For  two-phase  circuits.  —  Transformers  may  be  operated  on 
each  phase  of  a  two-phase  system,  all  the  connections  given  for 
single-phase  circuits  being  available.  In  addition,  the  primaries, 
or  the  secondaries,  or  both,  may  be  interconnected  as  explained  for 

the  polyphase  system  in  Chapter  7. 
4.  For  three-phase  circuits. — 
Single-phase  transformers  may 
be  connected  between  any  two 
lines  of  a  three-phase  system,  or 
between  any  line  and  the  neutral 


FIG.  1^6.     Delta-delta  Connection.  r  , 

of  a  star-connected  system,  and 

operated  as  from  a  single-phase  circuit.  The  primaries  of  three 
single-phase  transformers  may  be  connected  to  a  three-phase  supply 
system  in  either  star  or  delta;  the  secondaries  of  the  transformers 
may  be  connected  to  the  load  circuit  in  either  star  or  delta.  The 
voltage  and  current  relations  are  those  given  for  the  polyphase 
system  in  Chapter  7. 


FIG.  i57a.     Correct  Delta  Connections.     FIG.  1570.     Incorrect  Delta  Connections. 
Instantaneous  Voltages  Balanced.  Instantaneous  Voltages  Unbalanced. 

In  making  three-phase  transformer  connections  the  relations  in 
the  primary  coils  adjust  themselves.  If  the  secondaries  are  to  be 
connected  in  delta,  the  voltage  relations  must  be  as  indicated  in 
Fig.  i57a,  i.e.,  the  sum  of  the  instantaneous  electromotive  forces 
around  the  delta  must  be  zero.  If  any  one  of  the  coils  is  reversed 
the  equilibrium  of  voltages  is  destroyed,  and 
an  excessive  current  flows  around  the  delta. 
Fig.  isyb. 

If   the   secondaries   are    to   be    star-connected, 
FIG.  158.   Star-star          ,  .....  .,  ,     ,  .,  M      j 

Connection         a  snort  circuit  is  impossible  because  the  coils  do 

not  form  a  closed  circuit,  but  the  voltages  be- 
tween lines  may  be  unequal.  The  correct  relations  are  indicated 
in  Fig.  i5Qa,  and  the  voltage  between  any  two  lines  is  equal 
to  that  between  any  other  two  lines.  If  the  terminals  of  any 
one  of  the  coils  is  reversed,  the  relations  shown  in  Fig. 


TRANSFORMER  CONNECTIONS 


199 


exist  and  are  indicated  by  the  fact  that  the  voltage  between  two 
of  the  lines  is  correct,  while  that  between  either  of  these  and  the 
third  line  is  equal  to  the  voltage  in  the  coil,  i.e.,  to  the  voltage 
between  line  and  neutral. 


60' 


njr/ 


N 
c 

FIG.  i5Qa.    Correct  Star  Connections.      FIG.  i5gb.    Incorrect  Star  Connections. 
AB  =  AC  =  BC.  AB 

AL>  =  LJJ  =    — -=. 


V  or  open-delta  connection.  —  If  one  transformer  of  a  delta  connec- 
tion is  omitted,  a  three-phase  system  may  be  operated  with  only  two 
transformers.  This  is  the  V  or  open-delta  connection.  Fig.  161. 
While  the  number  of  transformers  required  for  this  connection  is 


FIG.  1 60.     Delta-star  (or  Star-delta) 
Connection. 


FIG.  161.    Open-delta  (or  V) 
Connection. 


oooooooooooo  P 


OGOGOOOOO 


less  than  in  the  delta  connection,  its  use  is  not  advised,  except  as 
a  temporary  expedient  for  the  continuation  of  service,  because  of 
the  phase  relations  of  the  current  and  the  electromotive  force  in 
the  windings  of  the  transformers.* 

5.  Phase  transformation.  —  It  is  some- 
times desirable  to  change  two-phase  cur- 
rent to  three-phase,  or  three-phase  to 
two-phase.  This  may  be  accomplished 
by  the  "T"  or  Scott  connection  in  which 
two  transformers  are  connected  as  indi- 
cated in  Fig.  162,  i.e.,  the  two-phase  coils 
are  independent  of  each  other  while  one  terminal  of  one  three- 
phase  coil  is  connected  to  the  middle  point  of  the  other  three- 
phase  coil.  The  terminals  A,  B  and  C  are  connected  to  the 
three-phase  lines. 

*  See  Chapter  7,  Section  30. 


FIG.  162.    Scott  (orT) 
Connection. 


200  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

The  voltage  of  the  coil  AC  is  that  between  the  lines  to 
which  it  is  connected.  The  voltage  of  the  coil  BD  is  a  component 

AC 

of  the  voltage  AB  (or  BC),  the  other  component  being  -  -  (  =  AD 

2 

=  DC).    Fig.  163. 

B  Since  ABC  is,  by  construction,  an  equilateral 

triangle  and  AD  =  DC,  ADB  is  a  right  triangle 
and 

V(ABy  -  (ADY  to 

0.866  (AB).  (2) 

FIG.     163.     Vector 

Diagram    of    T-       The  voltage  between  the  terminals  of  BD  is, 
connection.  therefore,  86.6  per  cent  of  the  voltage  between 

the  terminals  of  AC,  and  two  similar  transformers  do  not  induce 
equal  electromotive  forces  in  the  two-phase  windings  when  the 
transformers  are  supplied  from  a  balanced  three-phase  system. 
The  voltages  induced  in  the  two-phase  windings  are  equal  if  the 
ratio  of  the  number  of  turns  in  BD  to  the  number  of  turns  in 
AC  is  made  equal  to  0.866. 

No.  turns  in  coil  BD  _         ,  .  , 

No.  turns  in  coil  AC 

6.  For  synchronous  converters.  —  Transformer  connections  for 
two-,  three-  and  four-ring  converters  differ  in  no  way  from  those 
connections  already  described.  Because  of  the  increased  output 
of  a  given  converter  armature  when  provided  with  six  rings,  six- 
ring  converters  are  desirable.  For  supplying  current  to  six-ring 
converters,  the  primary  windings  of  transformers  may  be  con- 
nected either  star  or  delta  to  three-phase  mains,  but  the  secondary 
windings  must  be  connected  so  as  to  produce  six-phase  currents. 
These  connections  are:  (a)  diametral,  (6)  double  delta,  (c)  double 
star,  (d)  hexagonal. 

(a)  Diametral.  —  In    this    connection    the   terminals   of   trans- 
former A  are  connected  to  rings  i  and  4  of  the  converter,  the 
terminals  of  transformer  B  to  rings  2  and  5,  and  the  terminals  of 
transformer  C  to  rings  3  and  6.       Fig.  i64a. 

(b)  Double  delta.  —  The  windings  of  the  transformers  are  con- 
nected so  as  to  form  two  separate  deltas,  as  indicated  in  Fig.  i64b. 
The  parallel  sides  of  the  deltas  are  formed  by  the  coils  of  one  trans- 
former. 


TRANSFORMER   CONNECTIONS 


2OI 


(c)  Double  star.  —  The  six  coils  of  a  transformer  may  be  con- 
nected into  a  double  star,  as  indicated  in  Fig.  i64c.  In  this  con- 
nection the  coils  of  a  given  transformer  are  diametrically  opposed. 


000000000. 

.OOQOOOOOO 

poooooooo. 

0000000(50 

WOOWoW 

000000000 

A 

9 

c 

4  2       •'••        53  6 

FIG.  1643,.     Diametral. 


FIG.  i64b.    Double  Delta. 


FIG.  1640.    Double  Star. 


FIG.  1640!.    Hexagonal. 


(d)  Hexagonal.  —  The  six  coils  may  be  connected  as  indicated 
in  Fig.  i64d,  connection  being  made  to  the  rings  of  the  converter 
from  the  junction  of  each  two  coils.  As  in  the  double  delta,  par- 
allel sides  of  the  hexagon  are  formed  by  the  coils  of  one  transformer. 

From  the  voltage  relations  given  above  it  is  a  simple  matter  to 
make  any  of  the  above  connections  when  three  identical  trans- 
formers are  used.  When  the  transformers  are  not  similar,  the 
relative  direction  of  the  electromotive  forces  in  the  coils  must  be 
determined. 

CHAPTER  XII  — PROBLEMS 

1.  The  primaries  of  three  similar  20:1  transformers  are  star-connected  to 
23oo-volt  mains.     Find  the  voltage  of  the  secondary  circuit  when  the  second- 
aries are:  (a)  star-connected,  (6)  delta-connected. 

2.  Same  as  Problem  i  except  the  primaries  are  delta-connected. 

3.  A  5  kv-a.  transformer,  the  regulation  of  which  is  2  per  cent  is  operated  in 
parallel  with  another  5  kv-a.  transformer,  the  regulation  of  which  is  4  per  cent. 
When  the  total  load  on  the  circuit  is  equal  to  10  kv-a.  it  is  equally  divided  be- 
tween the  transformers.    Determine  the  load  division  when  the  total  load  is 
(a)  7.5  kv-a.,  (6)  5  kv-a.,  (c}  2.5  kv-a.    Assume  that  the  voltage  characteristics 
are  straight  lines. 


202  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

4.  Three   similar   transformers    have    their  primaries    star-connected    to 
balanced   three-phase  mains.     The  voltages  between  the  terminals  of  the 
star-connected  secondary  windings  are 

AB  =  127,  BC  =  127,  AC  =  220.  i 

State  what  the  trouble  is  and  how  it  may  be  remedied. 

5.  Two  100  kv-a.  transformers  are  V-connected.     Find  the  load  on  each 
transformer  when  the  balanced  load  on  the  system  is  equal  to  150  kw.,  and  the 
power  factor  is:  (a)  unity,  (b)  0.866. 

6.  The  primaries  of  two  similar  10:1  transformers  are  connected  to  2300- 
volt  2-phase  mains.     The  secondaries  are  T-connected.     Find  the  voltages 
between  the  3 -phase  (secondary)  terminals. 

7.  Draw  a  vector  diagram  showing  the  relations  of  the  current  and  the  elec- 
tromotive force  in  the  3 -phase  coils  of  a  T-connection  when  the  power  factor  of 
the  system  is  0.866  and  the  load  is  balanced. 

8.  The  primaries  of  three  similar  10  :  i  transformers  are  delta-connected  to 
23oo-volt  mains.    Determine  the  voltage  of  a  single-phase  circuit  when  the 
three  secondaries  are  connected  in  series. 

9.  A  transmission  line  is  connected  to  66oo-volt  generators  through  i :  10 
step-up  transformers  delta-connected  to  the  generators,  and  star-connected  to 
the  line.     Find  the  line-to-line  voltage. 

10.  A  3-phase  line  supplies  an  induction  motor  delivering  500  brake  horse 
power  through  a  bank  of  transformers.    The  primaries  are  star-connected;  the 
secondaries  are  delta-connected.    Ratio  60  :  i.    Voltage  between  lines  at  trans- 
former (primary)  terminals  =  45,000.     Transformer  efficiency  =  98.7  per  cent, 
motor  efficiency  =  92  per  cent,  power  factor  of  motor  =  87  per  cent.    Find:  (a) 
the  voltage  impressed  on  the  motor,  (V)  the  current  per  line  delivered  to  the 
motor,  (c)  the  kw.  supplied  to  the  transformers. 


CHAPTER  XIII 
THE   INDUCTION   MOTOR 

i.   Construction.  —  The  essential  parts  of  an  induction  motor 
are:  (a)  the  stator,  (b)  the  rotor. 

(a)  The  stator.  —  The  stator  of  an  induction  motor  is  the  station- 
ary part,  and  its  structure  is  essentially  that  of  the  armature  of  a 
rotating    field    alternator.      A 

slotted  core  (Fig.  166)  is  built 
up  of  laminations  (Fig.  167), 
and  the  stator  conductors, 
which  are  connected  to  the 
supply  mains,  are  placed  in  the 
slots  so  as  to  form  a  distributed 
winding. 

(b)  The     rotor.  —  Induction 
motors    are    differentiated    by 
the  construction  of  the  rotor, 
and  are:  (i)  squirrel  cage,  (2) 
slip-ring. 

(i)  Squirrel-cage  rotor. — A  squirrel-cage  rotor  consists  of  cop- 

, .    per   bars   or    rods   placed   in    slots 

on  the  •  surface  of  a  cylindrical 
laminated  iron  core,  the  ends  of  the 
conductors  being  connected  to  cop- 
per rings,  placed  at  each  end  of  the 
core.  Since  the  conductors  are 
connected  in  parallel,  the  resistance 
of  such  a  winding  is  small.  A 
squirrel-cage  rotor  is  shown  in  Fig. 
i68a. 

(2)  Slip-ring  rotor.  —  A  slip-ring 
or  wound  rotor  is  one  in  which 
distinct  windings,  similar  to  those  on  the  stator,  are  intercon- 
nected, and  terminals  brought  out  to  slip  rings  mounted  on  the 

203 


FIG.  165.    Induction  Motor  with  Wound 
Rotor.    Triumph  Electric  Co. 


FIG.  166.    Stator  Core  and  Frame. 
Crocker- Wheeler  Co. 


2O4 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


shaft.     Through  these  slip  rings  the  rotor  winding  is  connected  to 
an  external  rheostat,  by  means  of  which  the  resistance  of  the  rotor 

circuit  may  be  varied.  Since  the 
conductors  of  a  wound  rotor  are 
connected  in  series,  its  resistance 
is  much  greater  than  that  of  a 
squirrel  -  cage  rotor.  Fig.  i68b 
shows  a  wound  rotor  with  three 
rings  mounted  on  the  shaft. 

The  rheostat  for  regulating  the 
resistance  of  the  rotor  circuit  is 
sometimes  placed  in  the  rotor  struc- 
ture of  small  motors.  In  this 
case  no  rings  are  required,  but  a 
rod,  by  means  of  which  the  resist- 
ance of  the  rotor  circuit  is  changed,  projects  through  the  hollow 
shaft. 

2 .  Rotating  flux  and  its  produc- 
tion. —  It  was  shown  in  Chapter 
8,  Section  9,  that  the  magneto- 
motive force  due  to  the  rotating 
armature  of  a  two-phase  alterna- 
tor is  constant  in  value  and  fixed 
in  direction.  When  the  arma- 


FIG.  167.      Stator  Lamination. 
Crocker-Wheeler  Co. 


FIG.  i68a.     Squirrel-cage  Rotor. 
Triumph  Electric  Co. 


ture  is  stationary,  the  magneto- 
motive force  of  the  armature 
winding  is  constant  in  value,  but  is  momentarily  changing  in  direc- 
tion, and  the  flux  ro- 
tates at  a  speed  pro- 
portional to  the  fre- 
quency of  the  current 
flowing  in  the  arma- 
ture conductors. 

Consider    two    elec- 
tromagnets  placed   at 
FIG.  i68b.     Slip-ring  Rotor.     Crocker- Wheeler  Co.      right    angles     to     each 

other  as  shown  in  Fig.  169,  and  excited  from  a  two-phase  alter- 
nating-current system.  It  is  evident  that  the  flux  due  to  winding 
A  is  maximum  when  that  due  to  winding  B  is  zero.  As  flux  A 


THE  INDUCTION   MOTOR 


205 


decreases,  flux  B  increases.  During  the  quarter  cycle  required  for 
flux  A  to  decrease  to  zero  and  for  flux  B  to  increase  to  maximum, 
a  resultant  field  is  produced,  the  instantaneous  value  and  direc- 
tion of  which  is  represented  by  the  vector  sum  of  the  instanta- 


FIG.  169. 


FIG.  170. 


neous  values  of  the  quadrature  fluxes.*  The  flux  due  to  each 
winding  of  Fig.  169,  and  the  resultant  flux  is  represented  in  Fig. 
170  when  flux  A  is  maximum,  30,  60  and  90  degrees  later. 

It  is  evident  that  the  maximum  value  of  the  resultant  flux  is  con- 
stant and  equal  to  the  maximum  flux  produced  by  each  winding 
separately,  and  that  its  rate  of  rotation  is  proportional  to  the 
frequency  of  the  supply  circuit. 

It  may  be  shown  in  a  similar  manner  that  a  rotating  flux,  the 
maximum  value  of  which  is  1.5  times  the  maximum  flux  set  up  by 
each  phase  winding,  is  produced  by  connecting  properly  designed 
coils  to  a  three-phase  supply  system. 

The  magnetic  effect  when  the  stator  windings  of  a  polyphase  in- 
duction motor  are  connected  to  a  circuit  supplying  polyphase  alter- 
nating current,  is  essentially  that  described,  the  difference  being  due 
to  the  arrangement  of  the  windings,  which  is  purely  a  mechanical 
detail,  and  to  the  reaction  of  the  currents  set  up  in  the  conductors 
of  the  rotor. 

3.  Generator  action.  —  The  rotating  flux  set  up  by  the  stator 
windings  of  an  induction  motor  cuts  across  the  conductors  of  the 
rotor,  induces  in  them  an  electromotive  force,  and  causes  a  current 
to  flow  in  the  rotor  circuit. 

4.  Motor  action.  —  The  rotating  flux  set  up  by  the  stator  wind- 
ings reacts  with  the  currents  induced  in  the  rotor  conductors  to 
produce  a  torque,f  and  the  direction  of  the  torque  is  the  same  as 
the  direction  of  flux  rotation. 

*  The  fluxes  are  in  quadrature  as  regards  both  time  and  space  (direction), 
f  See  Chapter  2,  Section  14. 


206  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

In  a  two-phase  motor,  the  direction  of  flux  rotation  is  reversed 
by  reversing  the  terminal  connections  of  either  phase  winding;  in 
a  three-phase  motor,  the  direction  of  flux  rotation  is  reversed  by 
reversing  any  two  line  connections. 

5.  Speed  of  rotor.*-  -The  torque  produced  by  the  reaction  of 
the  stator  flux  and  the  rotor  currents  causes  the  rotor  to  revolve. 
Because  the  rate  at  which  the  flux  cuts  the  rotor  conductors  de- 
creases as  the  speed  of  the  rotor  increases,  both  the  induced 
electromotive  force  and  the  rotor  current  decrease,  and  the  rotor 
runs  at  a  speed  which  establishes  equilibrium  in  the  system.  If 
the  load  increases,  the  speed  decreases  until  the  reaction  between 
the  flux  and  the  increased  current  produces  the  required  torque; 
if  the  load  decreases,  the  speed  increases  until  the  current  decreases 
to  a  value  which  reestablishes  equilibrium  in  the  system. 

The  speed  of  the  rotor  can  never  attain  the  speed  of  the  rotating 
flux  because  at  this  speed  (the  synchronous  speed)  no  current  flows 
in  the  rotor  windings,  and  no  torque  is  exerted  to  compensate  for 
the  frictional  and  other  losses  of  the  motor. 

6.'  Slip  of  the  rotor.  —  The  difference  between  the  speed  of  the 
rotating  flux  and  that  of  the  rotor  is  termed  the  "slip"  of  the  rotor. 
The  slip  is  approximately  proportional  to  the  load  over  the  normal 
working  range  of  the  motor.  At  overloads  the  slip  increases  faster 
than  the  load  until  maximum  torque  is  reached  after  which  both 
speed  and  torque  decrease  very  rapidly. 

The  slip  of  an  induction  motor  is  easily  measured,  unless  it  be- 
comes excessive,  by  the  stroboscopic  method.  On  the  end  of  the 
shaft  or  the  pulley,  mark  as  many  equally-spaced  radial  lines  as  there 
are  pairs  of  poles  on  the  motor,  and  illuminate  these  lines  by  means 
of  an  arc  lamp  connected  to  the  circuit  from  which  the  motor  re- 
ceives its  current.  When  the  motor  is  in  operation,  the  radial  lines' 
appear  to  rotate  in  a  direction  opposite  to  that  of  the  rotor.  The 
speed  of  this  apparent  rotation  is  proportional  to  the  slip  of  the 
rotor. 

The  stroboscopic  method  of  slip  measurement  depends  on  the 
fact  that  the  light  from  an  alternating-current  arc  lamp  is  pulsa- 
ting. •  If  the  rotor  moved  at  synchronous  speed,  the  radial  lines 

*  The  induction  motor  is,  inherently,  an  approximately  constant-speed  machine. 
Large  variations  in  speed  are  obtained  only  by  a  sacrifice  of  efficiency  or  of  mechanical 
simplicity. 


THE  INDUCTION  MOTOR  207 

would  advance  through  the  angular  distance  of  one  pole  pitch 
for  each  pulsation  of  the  light,  and  successive  pulsations  would 
show  the  lines  in  the  same  relative  positions.  But  the  angular 
advance  of  the  lines  is  less  than  the  angular  pitch  of  the  poles, 
and  each  successive  pulsation  of  the  light  shows  the .  lines  in  a 
position  slightly  behind  that  which  it  occupied  at  the  previous 
pulsation. 

Example. — A  four-pole  induction  motor  is  operated  from  60- 
cycle  supply  mains.  Determine  the  slip  of  its  rotor  when  127 
radial  lines  pass  through  the  field  of  vision  in  one  minute, 


^—    -  =  1.76  per  cent. 
60  X  60  X  2 


Another  simple  method  for  the  determination  of  slip  is  to  connect 
a  contact  maker  to  the  shaft  of  the  motor  so  that  it  closes,  once  in 
each  revolution  of  the  rotor,  the  circuit  of  a  voltmeter  connected 
across  the  mains  supplying  the  motor.  The  voltmeter  pointer 
swings  back  and  forth,  the  rate  of  the  swing  being  proportional  to 
the  slip  of  the  motor.  If  an  electrodynamometer  type  of  volt- 
meter is  used,  the  rate  at  which  the  pointer  swings  is  twice  as  great 
as  if  the  voltmeter  is  of  the  permanent  magnet  type. 

7.  Torque.  —  From  the  above  it  might  be  supposed  that  the 
maximum  torque  of  an  induction  motor  is  exerted  at  zero  speed,  since 
both  the  induced  electromotive  force  and  the  rotor  currents  are  then 
maximum.  But  the  frequency  of  the  rotor  currents  is  proportional 
to  the  slip  of  the  rotor,  and  the  rotor  current,  therefore,  lags  behind 
the  induced  electromotive  force  by  a  constantly  increasing  angle  as 
the  slip  increases.  The  lagging  current  tends  to  set  up  a  flux  which 
is  opposed  to  that  set  up  by  the  stator  windings.*  When  the  slip 
becomes  large,  this  demagnetizing  action  is  excessive,  and  the  flux 
decreases  faster  than  the  rotor  current  increases.  Therefore  the 
speed-torque  curve  of  an  induction  motor  is  not  a  straight  line, 
but  has  the  general  shape,  for  a  rotor  circuit  of  constant  resist- 
ance, shown  in  Fig.  171. 

Starting  slightly  below  synchronous  speed  (the  speed  of  the  ro- 
tating flux),  the  torque  increases  as  the  speed  decreases  until  the 
point  of  maximum  torque  is  reached.  At  speeds  less  than  that 
at  which  maximum  torque  is  developed,  the  torque  decreases  very 

*  See  Chapter  8,  Section  9. 


208 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


rapidly.     Consequently,  when  an  induction  motor  is  loaded  beyond 
the  point  of  maximum  torque,  it  stops. 

An  examination  of  the  speed-torque  curve  of  an  induction  motor 
shows  that  the  characteristic,  Over  the  working  range  of  the  motor, 


I70 


JO 


10 


50  100  150          200         250         300 

PER  CENT  OF  FULL  LOAD(RATED)TORQUE 

FIG.  171.     Speed-torque  Characteristic  for  Squirrel-cage  Induction  Motor. 


is  similar  to  that  of  a  shunt  (continuous-current)  motor,  i.e.,  the 
speed  drops  slightly  as  the  load  increases.  When  the  applied  volt- 
age is  that  at  which  the  motor  is  rated,  the  maximum  torque  of 
an  induction  motor  is  usually  from  two  to  three  times  its  rated 
load  torque;  its  starting  torque  is  from  one  and  one-half  to  two 
times  rated  value,  and  the  starting  current  is  from  five  to  six 
times  that  at  rated  load.* 

The  relations  between  the  slip  and  the  torque  of  an  induction 
motor  may  be  derived  as  follows:  Let 

Er  =  the  electromotive  force  induced  in  the  rotor  circuit  at 

zero  speed, 

Ir  —  the  rotor  current, 
Rr  =  the  resistance  of  the  rotor  circuit, 
Xr  =  the  reactance  of  the  rotor  circuit  at  zero  speed, 
s  =  the  slip  of  the  rotor  expressed  as  a  fraction  of  the 

synchronous  speed, 

*  The  Standardization  Rules  of  the  American  Institute  of  Electrical  Engineers 
require  that  a  motor  intended  for  continuous  service  shall  develop  a  maximum  run- 
ning torque  at  least  75%  greater  than  the  normal  torque  at  rated  load. 


THE  INDUCTION  MOTOR 


209 


But 
and 


n'  =  the  synchronous  speed  of  the  rotor  in  revolutions  per 

minute, 

n"  =  the  actual  speed  of  the  rotor  in  revolutions  per  minute, 
cos  (f>  =  the  power  factor  of  the  rotor  circuit. 

Rotor  input  =  Erlr  cos  <£.  (i) 


sEr 


100 

80 
70 

1, 

COS  <f> 
s?55^ 

VRS  +  s2xr2' 

Rr 

«~- 

•^ 
•^^ 

\ 

—  . 

'A 

~—  - 

^ 

^ 

—  * 

2 

M« 
—  . 

4 

^~— 
«—  , 

V1-' 

T 
01 

x2 

M  1  1  1  1  1  1  1  1  1  1  1 

f 

^-^ 

*  —  , 

'~—  1 

V 

s^ 

x 

X 

•* 

K 

^ 

K^ 

\ 

^ 

X, 

^ 

^ 

"•^ 

1  —  , 

Pi 

^-> 

^ 

^ 

i^ 

s 

i^ 

^ 

S^ 

*•» 

^ 

^ 

> 

x 

^ 

\ 

^ 

N 

^ 

s 

•>> 

4> 

*** 

ix^ 

s 

k^ 

\^ 

X, 

*x 

•^ 

s 

s, 

^ 

^ 

N 

x 

kv 

s 

SN 

^ 

s, 

\ 

s^ 

x, 

Xj 

^ 

V 

S 

S, 

^ 

k^ 

s 

V  ' 

s^ 

s 

^v 

Xj 

y1 

\ 

s, 

s 

f 

i 

\ 

Trt 

\yn 

s, 

f 

\ 

s. 

\ 

s 

/ 

^ 

/ 

7O 

s^ 

' 

^ 

S 

^ 

\ 

k^ 

\ 

} 

*\ 

1 

10 

s 

s 

j 

s 

/ 

t 

^ 

/ 

\ 

^ 

ss 

1 

/ 

\ 

/ 

/ 

y 

(3) 


SO  100          \50         200          150         300 

PERCENT  OF  FULL  LOAD  (RATED)  TORQUE 

FIG.  172.     Speed-torque  Characteristics  for  Slip-ring  Induction  Motor.     Resistance 
of  Rotor  Increased  by  Steps. 


Substituting  in  equation  (i), 
Rotor  input  = 

Rotor  output  = 


watts, 


=  sEr2Rr  (i  -  s) 
R2  +  s2X2 


Rr2  +  S2X2 

watts, 


Torque 


33,000 sE2Rr(i  -s) 


n'(R2+s2X2) 


r-^-  foot-pounds. 


(4) 
(S) 
(6) 
(7) 
(8) 
(9) 


210  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

The  above  formulae  disclose  the  following  operating  character- 
istics of  the  induction  motor: 

(a)  Maximum  torque  is  developed  when 

Rr  =  sXr  (10) 

(b)  Maximum  torque 


and  is  independent  of  rotor  resistance. 

(c)  Starting  torque 

7-°4Er2Rr  (       x 

-n'(R*  +  X?Y 
is  proportional  to  the  rotor  copper  losses,  and  is  maximum  when 

Rr  =  Xr.  (13) 

(d)  The  copper  losses  in  the  rotor  circuit  are  equal  to  the  product 
of  the  rotor  input  and  the  slip. 

(e)  At  speeds  near  synchronism,  the  reactance  of  the  rotor  cir- 
cuit is  negligible,  the  torque  is  directly  proportional  to  the  slip, 
inversely  proportional  to  the  resistance  of  the  rotor  circuit,  and 


(/)  At  maximum  torque,  the  power  factor  of  the  rotor  circuit 

=  0.707.  (15) 

(g)  At  constant  speed,  the  torque  is  directly  proportional  to  the 
square  of  the  electromotive  force  induced  in  the  rotor  circuit  at 
zero  speed,  and,  therefore,  to  the  square  of  the  applied  voltage. 

(ti)  At  constant  torque,  the  slip  of  the  rotor  is  inversely  propor- 
tional to  the  square  of  the  electromotive  force  induced  in  the  rotor 
circuit  at  zero  speed,  and,  therefore,  to  the  square  of  the  applied 
voltage. 

8.  Starting.  —  The  polyphase  induction  motor  is  self-starting 
when  supplied  with  alternating  current  of  the  proper  voltage,  fre- 
quency, and  number  of  phases.  The  squirrel-cage  motor  is  usually 
started  by  supplying  the  stator  with  current  at  a  voltage  less  than 
that  at  which  the  motor  is  rated,  the  line  voltage  being  reduced  by 
means  of  an  auto-transformer  or  other  step-down  device.  Fig.  173. 
After  the  rotor  has  attained  a  considerable  speed,  the  line  voltage  is 


THE  INDUCTION  MOTOR 


211 


^Compensator 
Auto-transformer) 


applied  and  the  starting  device  automatically  disconnected  from  the 
line.  Starting  at  a  reduced  voltage  minimizes  the  line  disturbances 
and  reduces  the  heating  of  the  motor  windings,  but  decreases  the 
starting  torque. 

The  slip-ring  induction  motor  is  started  by  applying  rated  volt- 
age to  the  stator  windings,  after  resistance  has  been  inserted  in  the 
rotor  circuit.  As  •  the  rotor  speeds  up,  the 
resistance  of  the  rotor  circuit  is  gradually 
reduced  until  the  windings  are  short-circuited. 

Because  of  their  large  starting  currents  and 
low  power  factors,  induction  motors  often  cause 
undesirable  fluctuations  of  the  line  voltage  dur- 
ing the  starting  period.  These  fluctuations 
are  particularly  objectionable  when  lamps  and 
motors  are  operated  in  parallel. 

9.  Power  factor.  —  Since  the  air  gap  of  an 
induction  motor  introduces  considerable  reluc- 
tance into  the  magnetic  circuit,  the  mag- 
netizing current,  as  compared  to  the  load  FlG<  I73'  ComPensat<* 

.     .     .     .  Connections, 

current,  is  relatively  large  at  small  percentages 

of  the  rated  load.  The  power  factor  of  an  induction  motor  is, 
therefore,  low  at  no-load  but  increases  as  the  load  increases. 

The  frequency  of  the  currents  induced  in  the  rotor  circuit  is  pro- 
portional to  the  slip  of  the  rotor.  The  reactance  of  the  rotor  circuit 
is,  then,  proportional  to  the  slip  of  the  rotor. 

"V  f  fT  /     /-\ 

Ar     =2  irSjLy  (lO) 

when   XT'  =  the  reactance  of  the  rotor  circuit,  at  slip  s, 

s  =  the  slip  of  the  rotor  expressed  as  a  fraction  of  the 

synchronous  speed, 

/  =  the  frequency  of  the  supply  circuit, 
L  =  the  inductance  of  the  rotor  circuit. 

When  the  slip  is  small  the  reactance  of  the  rotor  circuit  is  neg- 
ligible, the  impedance  of  the  circuit  is  practically  equal  to  its 
resistance,  and  the  rotor  current  is  in  phase  with  the  induced  elec- 
tromotive force.  As  the  slip  increases,  the  reactance  of  the  rotor 
circuit  increases,  the  impedance  becomes  greater  than  the  resistance, 
and  the  current  lags  behind  the  induced  electromotive  force. 

The  power  factor  of  an  induction  motor  will  increase  as  long  as 


212  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

the  power  component  of  the  induced  voltage  increases  faster  than 
the  reactive  component.  When  the  slip  becomes  so  great  that  the 
reactive  component  of  the  voltage  increases  faster  than  the  power 
component,  the  power  factor  decreases. 

10.  The  losses.*  —  The  losses  in  an  induction  motor  are:  (a) 
copper  losses,  (b)  stray  power. 

(a)  Copper  losses.  —  The  copper  losses  of  an  induction  motor  are 
those  due  to  the  resistance  of  the  stator  windings  and  that  of  the 
rotor  circuit.  The  losses  in  the  stator  windings  are  easily  calculated 
for  any  given  value  of  stator  current  when  the  resistance  of  the 
windings  is  known.  The  resistance  is  determined  by  continuous- 
current  methods. 

The  copper  losses  in  a  slip-ring  rotor  are  as  easily  determined  as 
are  those  of  the  stator.  Those  in  a  squirrel-cage  rotor  cannot  be 
determined  directly,  since  neither  the  resistance  of  the  circuit  nor 
the  value  of  the  current  flowing  in  it  can  be  measured.  The  rotor 
copper  loss  should  be  calculated  by  means  of  the  following  formula: 

„  T  g      output  X  slip  ,v 

Krlr  "         i  -  slip      '  U7J 

Let 

Er  =  the  electromotive  force  induced  in  the  rotor  circuit  at 

zero  speed, 

Rr  =  the  resistance  of  the  rotor  circuit, 
Xr  =  the  reactance  of  the  rotor  circuit  at  zero  speed, 
s  =  the. slip  of  the  rotor  expressed  as  a  fraction  of  the  syn- 
chronous speed, 
7r  =  the  current  in  the  rotor  circuit  at  slip  s. 

,       induced  e.m.f. 

lr  =  —. — 

impedance 
sEr 


(19) 
Squaring  equation  (19)  and  multiplying  by  Rr 

(*>) 


=  ks.  (21) 

*  See  the  Standardization  Rules  of  the  American  Institute  of  Electrical  Engineers. 


THE   INDUCTION  MOTOR 


2I3 


(b)  Stray  power.  —  The  stray  power  of  an  induction  motor  in- 
cludes windage,  friction,  and  iron  losses,  and  is  approximately  con- 
stant over  the  working  range  of  speeds.  It  is  evident  that  windage, 
friction,  and  stator  iron  losses  decrease  as  the  slip  (load)  increases, 
and  that  the  iron  losses  in  the  rotor  increase.  The  stray  power  of 
an  induction  motor  is  usually  taken  as  the  no-load  input  to  the 


180 


25  50  75  100          125  150 

PERCENT  OF  RATED  OUTPUT (H.R) 

FIG.  174.    Induction  Motor  Performance  Curves. 


ZOO 


motor  minus  the  stator  copper  losses,  the  rotor  copper  losses  at  no- 
load  being  so  small  as  to  be  negligible. 

Stray  power  =  no-load  input  —  stator  RP.  (22) 

ii.  Performance  curves.  —  The  performance  of  an  induction 
motor  is  indicated  by  curves  such  as  those  shown  in  Fig.  174  data 
for  which  may  be  derived  from:  (a)  a  brake  test,  (b)  the  losses, 
(c)  the  circle  diagram. 

(a)  The  brake  test.  —  A  brake  test  is  usually  unsatisfactory  be- 
cause of  the  large  number  of  readings  required,  and  the  difficulty 


214 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


experienced  in  holding  a  brake  load  constant.     The  brake  test  is, 
therefore,  little  used  in  induction  motor  testing. 

(b)  The  losses.  —  The  set-up  for  the  determination  of  the  losses  is 
the  same  as  that  used  in  a  brake  test,  but  the  load  is  applied 
by  means  of  an  electric  generator  or  a  blower,  and  no  measure- 
ments of  output  need  be  taken.     The  stray  power  is  calculated 
from  the  no-load  input  and  the  stator  copper  losses  as  indicated 
above. 

Output  =  input  —  stray  power  —  stator  RP  —  rotor  RP.  (23) 

But  from  equation  (21)  the  rotor  copper  losses  are  proportional  to  the 
slip.     Hence, 

Output  =  (input  —  stray  power  —  stator  RP)  (1  —  5).     (24) 

(c)  The  circle  diagram.  —  It  may  be  proved  both  experimentally 
and  mathematically,  that  the  locus  of  the  vector  of  the  current  in  the 
rotor  circuit  is  a  semi-circle,  i.e.,  the  value  of  the  rotor  current  and 


FIG.  175.     Circle  Diagram  for  Induction  Motor. 

the  power  factor  of  the  circuit  are  so  related  that,  as  the  current 
varies,  the  end  of  the  vector  is  always  on  a  semi-circle  as  indi- 
cated in  Fig.  175.  Therefore,  the  semi-circle  is  determined  when 
two  points  are  determined,  the  diameter  being  parallel  to  the  axis 
of  abscissa.  The  two  points  usually  selected  for  experimental  deter- 
mination are:  (i)  when  the  motor  is  running  without  load  (no-load 
test),  (2)  when  the  rotor  is  blocked  to  prevent  its  rotation  (blocked 
rotor  test). 

(i)  No-load  test.  —  The  motor  is  supplied  with  current  at  rated 
voltage  and  run  without  load.  Measure  the  current  and  the  watts 
input  to  the  stator. 


THE   INDUCTION   MOTOR 


215 


t/3  <O 


(2)  Blocked  rotor  test.  —  After  blocking  the  rotor  to  prevent  its 
rotation,  apply  rated  voltage  to  the  stator,  and  determine  the  stator 
current  and  the  watts  input. 

The  excessive  currents  which  flow  in  the  windings  of  the  motor 
when  rated  voltage  is  applied  with  the  rotor  blocked,  may  be  avoided 
by  making  the  blocked  rotor  test  as  follows:  Wattmeter  and  ammeter 
readings  are  taken  for  several  values  of  applied  voltage  less  than 
that  at  which  the  motor  is  rated.  (Very  low  values  should  not  be 
used  as  the  results  are  likely  to  be  erratic.  If  the  stator  cur- 
rents do  not  exceed  twice  the  rated 
full-load  value,  no  damage  can 
be  done  to  the  windings  or  to  the 
insulation.)  A  curve  plotted  be- 
tween the  power  component  of  the 

,  /watts\ 

stator   current  -  and   volts 

V  volts/ 

is  a  straight  line  that  may  be  ex- 
tended to  any  desired  point,  the 
product  of  the  ordinate  and  the 
abscissa  of  the  point  being  equal 
to  the  watts  input  at  this  voltage. 
Fig.  176. 

12.  Construction  of  the  circle  diagram.  —  Draw  OA  (Fig.  175) 
proportional  to  the  rated  electromotive  force  of  the  motor,  using 
any  suitable  scale.  From  O  lay  off  OB  proportional  to  the  no-load 
current,  the  angle  AOB  being  made  such  that  its  cosine  is  equal  to 
the  power  factor  of  the  motor  at  no  load.  Also  lay  off  the  line  OC 
proportional  to  the  stator  current  with  the  rotor  blocked,  the  cosine 
of  the  angle  AOC  being  equal  to  the  power  factor  of  the  motor 
under  this  condition.  Draw  the  line  BD  parallel  to  the  axis  of 
abscissa.  The  points  B  and  C  are  two  points  on  the  semicircular 
locus  of  the  current  vector,  the  center  of  the  circle  being  on  the 
line  BD. 

The  perpendicular  from  C  to  the  axis  of  abscissa  is  proportional 
to  the  total  losses  in  the  motor.  Since  the  rotor  is  blocked  and  can 
deliver  no  mechanical  power,  this  loss  is  equal  to  the  input.  De- 
termine the  increase  in  the  stator  copper  loss  over  that  at  no-load, 
and  lay  off  as  EF. 

EF  =  R.  (722  -  A2),  (25) 


40  60  60  100 

PER  CENT  OF  RATED  VOLTAGE 

FIG.  176. 


2l6  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

when  R8  =  the  resistance  of  the  stator  windings, 

I2  =  the  stator  current  with  the  rotor  blocked, 
/i  =  the  stator  current  at  no-load. 

CF  is  proportional  to  the  copper  and  iron  losses  in  the  rotor.  No 
attempt  need  be  made  to  separate  these  losses  since  the  rotor  iron 
losses,  under  load  conditions,  are  negligible.  (The  frequency  of  the 
magnetic  reversals  in  the  rotor  iron  is  very  low.) 

Draw  straight  lines  from  C  and  F  to  B.  For  any  current  input. 
asOG, 

HK  is  proportional  to  the  no-load  (constant)  losses, 

KL  is  proportional  to  the  "added"  stator  copper  loss, 

LM  is  proportional  to  the  rotor  copper  losses, 

MG  is  proportional  to  the  output  of  the  motor, 

LG  is  proportional  to  the  torque  developed  in  the  rotor, 

TI  r  f~i 

*  is  the  ratio  of  the  rotor  speed  to  the  synchronous  speed, 
LG 

is  the  ratio  of  the  slip  to  the  synchronous  speed, 

LG 

~\ff^ 

— —  is  the  efficiency  of  the  motor, 

HG 

cos  of  the  angle  AOG  is  the  power  factor  of  the  motor  circuit. 

The  maximum  power  factor  at  which  an  induction  motor  can 
operate  is  obtained  when  the  current  vector  is  tangent  to  the  circle. 

The  maximum  output  of  an  induction  motor  is  obtained  when  the 
current  is  of  such  a  value  that  a  line  drawn  through  the  end  of  the 
current  vector,  and  tangent  to  the  circle,  is  parallel  to  EC. 

The  maximum  torque  possible  to  develop  in  an  induction  motor 
is  exerted  when  the  current  is  of  such  a  value  that  a  line  drawn 
through  the  end  of  the  current  vector,  and  tangent  to  the  circle,  is 
parallel  to  BF. 

It  will  be  observed  that  the  above  quantities,  as  determined  by 
the  circle  diagram,  are  not  absolutely  correct,  but  that  the  errors  are 
small  and  that  they  tend  to  neutralize  each  other.  For  example, 
the  so-called  constant  losses  decrease  slightly  as  the  slip  increases 
(windage  and  friction  are  dependent  on  the  speed  of  the  rotor 
and  are  zero  when  the  rotor  is  blocked),  while  the  rotor  iron  losses 
increase. 


THE  INDUCTION  MOTOR  217 

The  circle  diagram  offers  a  simple  method  for  determining  the 
actions  of  an  induction  motor;  the  results  are  sufficiently  accurate 
for  commercial  purposes;  and  the  only  data  required  are  the  current 
and  watts  input  at  no  load,  the  current  and  watts  input  with  the 
rotor  blocked,  and  the  resistance  of  the  stator  windings.* 

13.   Proof  of  the  circle  diagram.  —  Let 

Er  =  the  electromotive  force  induced  in  the  rotor  circuit  at 

zero  speed, 

Rr  =  the  resistance  of  the  rotor  circuit, 
Xr  =  the  reactance  of  the  rotor  circuit  at  zero  speed, 
s  =  the  slip  of  the  rotor  expressed  as  a  fraction  of  the  syn- 

chronous speed, 

7r  =  the  current  in  the  rotor  circuit  at  slip  s, 
<j>  =  the  phase  angle  between  Er  and  7r. 

Then 

.  (26) 


But  r  =  tan^  (27) 

Kr 

and  Rr  =  -^-  (28) 

tan<*> 

=   sXr  COt  0.  (29) 

Substituting  the  value  of  Rr  from  equation  (29)  in  equation  (26), 

*      T  sEr  /    \ 

Ir   =   —•=  (30) 


$XrV(*  - 

=  ^-  .';,.".  (32) 

Equation  (32)  is  a  polar  equation  of  the  circle.f 

*  In  plotting  the  circle  diagram  for  a  polyphase  motor  it  is  convenient  to  use  the 
quantities  for  one  phase,  in  which  case  the  watts,  torque,  output,  losses,  etc.,  must 
be  multiplied  by  the  number  of  phases  to  obtain  the  quantities  for  the  motor,  or  to 
reduce  the  experimental  quantities  to  the  "equivalent  single-phase"  system,  as  ex- 
plained in  Chapter  7,  Section  n. 

t  The  proof  given  above  is  strictly  true  only  when  the  resistance  of  the  stator  wind- 
ings and  the  magnetic  leakage  are  negligible,  but  over  the  operating  range  of  an  in- 
duction motor  very  small  differences  are  found  between  the  theoretical  and  the  actual 
locus  of  the  current  vector. 


218 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


THE  SINGLE-PHASE  INDUCTION  MOTOR* 

Structurally,  the  single-phase  induction  motor  is  essentially  the 
same  as  the  polyphase  motor,  but  the  stator  windings  are  connected 
to  a  single-phase  supply  system.  The  chief  operating  difference 
between  the  single-phase  and  the  polyphase  motor  is  the  fact  that 
the  former,  is  not  inherently  self-starting. 

14.  Transformer  action.  — When  the  windings  A  A  (Fig.  lyya) 
are  connected  to  a  single-phase  alternating-current  circuit,  the  flux 
set  up  by  the  windings  periodically  reverses  in  direction,  and  the 
changing  value  of  flux  induces  an  electromotive  force  in  the  conduc- 
tors of  the  squirrel-cage  structure.  But  the  directions  of  the  cur- 


FIG. 


FIG.  i;7b. 


rents  in  the  conductors  are  such  that  the  torque  exerted  on  one 
conductor  is  equal  and  opposite  to  that  exerted  on  another 
conductor.  There  is,  therefore,  no  tendency  for  the  conductors 
to  rotate,  the  effect  of  the  rotor  currents  being  to  largely  neutralize 
the  flux  set  up  by  the  windings  A  A,  the  action  being  identical  with 
that  in  a  transformer,  the  secondary  of  which  is  short-circuited. 

15.  Generator  action. —  If  the  structure  supporting  the  rotor 
conductors  is  rotated  by  some  external  power,  an  electromotive 
force  is  generated  in  the  conductors  by  reason  of  their  movement 
across  the  magnetic  field  set  up  by  windings  AA.\ 

The  current-carrying  conductors  on  the  rotor  structure  set  up 
a  quadrature  flux,  as  indicated  in  Fig.  lyyb.  During  the  half- 
cycle  in  which  the  current  in  AA  flows  in  the  direction  indicated  by 
the  arrows,  the  flux  set  up  by  A  A  is  constant  in  direction,  though 

*  For  a  mathematical  discussion  of  the  single-phase  induction  motor,  the  student 
is  referred  to  "  Electric  Motors  "  by  Crocker  &  Arendt. 
t  See  Chapter  2,  Section  13. 


THE  INDUCTION   MOTOR  219 

varying  in  magnitude,  and  the  direction  of  the  flux  set  up  by  the 
rotor  winding  is  that  indicated  m  the  figure.  During  the  next 
half-cycle  the  flux  due  to  the  windings  AA  is  reversed,  since  the 
current  direction  in  the  windings  is  reversed.  The  reversed  direc- 
tion of  the  field  across  which  the  rotor  conductors  move,  reverses 
the  direction  of  the  rotor  flux.  Therefore,  the  flux  set  up  by  the 
rotor  winding  of  a  single-phase  induction  motor  is  in  quadrature, 
both  in  time  and  in  space,  with  the  flux  set  up  by  the  stator 
windings,  and  alternates  in  direction,  the  frequency  of  the  flux 
reversal  being  the  same  as  that  of  the  current  reversal  in  the 
windings  AA.* 

It  is  evident  that  the  generated  electromotive  force  and,  there- 
fore, the  quadrature  flux,  is  proportional  to  the  speed  of  the  rotor; 
and  that  the  magnetomotive  force  of  the  rotor,  when  rotating  at 
synchronous  speed,  is  equal  to  the  magnetomotive  force  of  the  stator 
windings. 

1 6.  Rotating  flux.  —  While  the  simultaneous  existence  of  quad- 
rature magnetic  fluxes  in  the  same  material  is  entirely  imaginary, 
the  effects  are  real  and  are  due  to  the  actual  flux 

resulting  from  quadrature  components.  There 
is,  then,  set  up  in  a  single-phase  induction  motor, 
when  once  started,  a  rotating  flux  the  value  of 
which  is  constant  only  when  the  rotor  revolves 
at  synchronous  speed.  Since  no  electromotive 
force  could  be  induced  in  the  rotor  conductors  at 
synchronous  speed  (field  and  conductors  rotating  FlG  g  '  Rotatino. 
at  the  same  speed),  the  operating  speed  of  the  Flux  of  Single- 
rotor  is  always  less  than  synchronous  speed,  and  phase  Induction 
the  rotating  field  of  a  single-phase  induction  motor  Mo1 
is  an  ellipse,  as  indicated  in  Fig.  178,  the  short  axis  being  proportional 
to  the  speed  of  the  rotor. 

17.  Starting. — To  make  single-phase  induction  motors  com- 
mercially practicable,  auxiliary  starting  devices  are  required.     Two 
methods  of  producing  a  starting  torque  in  single-phase  induction 
motors  are  in  general  use,  and  will  be  described:  (a)  shading  coils, 
(b)  split-phase  windings. 

*  Because  of  the  reversal  of  the  quadrature  flux,  and  the  passage  of  the  rotor  con- 
ductors across  the  quadrature  field,  the  currents  flowing  in  the  rotor  conductors  of  a 
single-phase  induction  motor,  are  due  to  the  combined  action  of  four  electromotive 
forces. 


22O 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


(a)  Shading  coils.  —  A  shading  coil  is  a  closed  winding  placed 
around  a  portion  of  the  pole,  as  indicated  in  Fig.  179.  As  the  flux 
in  the  pole  alternates,  by  reason  of  the  reversal  of  the  electromotive 
force  of  the  supply  circuit,  an  electromotive  force  is  induced  in  the 

shading  coil.  The  effect  of  the  induced 
electromotive  force  is  to  oppose  any  change 
in  the  magnetic  conditions  existing  in  the 
space  enclosed  by  the  coil.  This  opposi- 
tion causes  the  flux  threading  the  coil  to  lag 
behind  the  flux  in  the  unshaded  part  of  the 
pole,  and  thus  reach  its  maximum  value 
at  a  later  period.  A  flux  which  moves 
(shifts)  from  the  unshaded  to  the  shaded 
part  of  the  pole  is  thus  produced  ,  and  a 
small  starting  torque  obtained.  Shading 

FIG.  179.    Single-phase  In-  coils  are  commonly  used  in  fan  and  other 
duction  Motor. 


(b)  Split-phase  windings.  —  When  a  single-phase  induction  motor 
is  to  be  started  by  the  split-phase  method,  an  auxiliary  stator  wind- 
ing must  be  provided  and  the  motor  is,  structurally,  a  polyphase 
motor.  The  connections  for  starting  a  two-phase  motor  from  a 


S*ifct> 

Running 
Starting 


Resistance 


FIG.  i8oa.    Split-phase  (Two-phase 
Winding). 


FIG.  i Sob. 


Split-phase  (Three-phase 
Winding). 


single-phase  circuit  are  shown  in  Fig.  i8oa,  and  those  for  a  three- 
phase  motor  in  Fig.  i8ob. 

Fig.  181  is  a  vector  diagram,  with  stationary  rotor,  of  the  circuits 
represented  in  Fig.  i8oa.  The  impedance  of  circuit  A  is  greater 
than  that  of  circuit  B  because  of  the  resistance  connected  in  series 


THE  INDUCTION  MOTOR 


221 


with  the  motor  winding;  the  power  factors  of  the  two  circuits  are 
different,  and  the  currents  are  out  of  phase  as  indicated..  The 
fluxes  set  up  by  the  windings  are,  therefore,  out 
of  phase,  and  produce  a  resultant  rotating  (or 
shifting)  field. 

It  is  impracticable  to  produce  a  quadra- 
ture displacement  of  fluxes  by  means  of  split- 
phase  connections,  but  a  displacement  sufficient 
to  produce  a  small  starting  torque  is  easily 
obtained.  The  direction  of  rotation  is  reversed  FIG.  181.  Vector  Dia- 
by  reversing  the  terminal  connections  of  either  gram  SpUt "  Phase 

.      .  Motor. 

winding. 

18.  Speed-torque  curves.  —  Fig.  182  shows  the  speed-torque 
relations  of  a  single-phase  induction  motor  with  variable  rotor 
resistance.  Both  the  maximum  torque  and  the  speed  at  which  it 


. 


\n 


50          KX)  150          200         250 

PERCENT  OF  FULL  LOAD  (RATED) 
TORQUE 

FIG.  182.    Speed- torque  Characteristics  of  Single-phase  Induction  Motor 
with  Variable  Rotor  Resistance. 

is  developed,  are  reduced  by  increasing  the  resistance  of  the  rotor 
circuit. 

THE  FREQUENCY  CHANGER 

19.  The  frequency  of  a  transmission  line  is  often  less  than  that 
required  for  circuits  supplying  lamps.  To  operate  lamps  from  such 
a  system  it  is  necessary  to  raise  the  frequency.  As  indicated  in 
Section  9,  the  frequency  of  the  current  in  the  rotor  circuit  of  an  in- 
duction motor  is  proportional  to  the  slip  of  the  rotor.  If,  then,  the 


222  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

slip-ring  rotor  of  an  induction  motor  is  driven  at  the  proper  speed 
a  current  of  any  desired  frequency  may  be  obtained  from  the  rotor 
windings.  When  so  used  the  induction  motor  is  termed  a  frequency 
changer.* 

The  voltage  at  the  terminals  of  the  rotor  circuit  is  proportional 
to  the  slip  of  the  rotor.  This  is  easily  demonstrated  by  considering 
the  voltage  at  synchronous  speed  and  at  zero  speed.  At  synchro- 
nous speed,  the  voltage  must  be  zero  since  there  is  no  relative  move- 
ment between  the  flux  and  the  rotor  conductors.  At  zero  speed,  the 
stator  and  the  rotor  act  as  the  primary  and  the  secondary  of  a  trans- 
former, and  the  voltages  are  to  each  other  as  the  number  of  turns 
in  the  windings.  The  voltage  of  the  rotor  circuit  at  any  given  speed 
is,  then,  obtained  by  multiplying  the  voltage  of  the  supply  circuit 
by  the  ratio  of  the  number  of  turns  in  the  windings  and  by  the  slip 
of  the  rotor  from  synchronism. 

Er  =  fo£.,t  (33) 

when  Er  =  the  electromotive  force  induced  in  the  rotor  winding, 
k  =  the  ratio  of  the  number  of  turns  on  the  rotor  to  the 

number  of  turns  on  the  stator, 
s  =  the  slip  of  the  rotor  expressed  as  a  fraction  of  the 

synchronous  speed, 
Es  =  the  voltage  of  the  supply  circuit  at  the  stator  terminals. 

The  ratio  k  of  the  number  of  turns  in  the  windings  may  be  deter- 
mined by  voltage  measurements  at  100  per  cent  slip  (standstill). 

The  rotor  of  an  induction  motor,  when  used  as  a  frequency 
changer,  is  driven  by  another  motor  (either  induction  or  synchro- 
nous) supplied  from  the  same  system  as  is  the  stator  of  the  frequency 
changer.  In  case  the  driving  motor  is  synchronous,  its  fields  may 
be  over  excited  to  compensate  for  the  lagging  current  of  the  fre- 
quency changer,  and  the  power  factor  of  the  supply  circuit  kept 
at  a  high  value. 

At  zero  speed,  it  is  evident  that  no  power  is  required  from  the 
driving  motor,  and  that  a  certain  electromotive  force  is  induced  in 
the  rotor  circuit  by  transformer  action.  If  the  rotor  is  driven  at 

*  The  required  change  in  frequency  may  be  affected  by  means  of  a  motor-generator 
set,  e.g.,  a  twenty-five  cycle  motor  (synchronous  or  induction)  driving  a  sixty-cycle 
alternator. 

t  This  equation  is  only  approximate  when  current  flows  in  the  windings,  because 
of  magnetic  leakage  and  stator  resistance. 


THE  INDUCTION  MOTOR  223 

synchronous  speed  backwards,  and  the  current  maintained  at  a  con- 
stant value,  the  voltage  of  the  circuit  is  doubled.  But  the  in- 
creased voltage  is  due  to  the  actual  movement  of  the  rotor  (i.e., 
to  generator  action),  and  the  increased  power  is  supplied  through 
the  motor. 

Since  there  is  no  fixed  relation  between  the  number  of  phases 
supplied  to  the  stator  and  the  number  for  which  the  rotor  may  be 
wound,  the  induction  machine  may  be  used  to  change  the  number 
of  phases  as  well  as  the  frequency  of  a  system. 

The  chief  objections  to  the  induction  frequency  or  phase  changer 
are  its  poor  regulation  and  low  power  factor,  due  to  the  large  air- 
gap  leakage  reactance. 

THE  INDUCTION  GENERATOR 

20.  When  the  rotor  of  an  induction  motor  is  driven  above  syn- 
chronism, the  counter-electromotive  force  of  the  motor  becomes 
greater  than  the  applied  electromotive  force,  and  the  power  compo- 
nent of  current  in  the  supply  circuit  is  reversed,  i.e.,  the  motor  acts 
as  a  generator  and  delivers  current  to  the  supply  system.  When 
so  operated,  an  induction  machine  is  not  self-exciting,  and  must  be 
operated  in  connection  with  synchronous  apparatus  from  which 
it  may  receive  its  magnetizing  current. 

If  an  induction  motor,  the  rotor  of  which  is  driven  above  syn- 
chronism by  an  independent  prime  mover,  is  connected  to  a  system 
supplied  by  an  alternating-current  generator  (synchronous  alter- 
nator), the  following  effects  are  noted  when  the  driving  torque  sup- 
plied to  the  alternator  is  reduced  to  zero: 

(a)  The  alternator  operates  as  a  synchronous  motor. 

(b)  If  the  field  excitation  of  the  synchronous  machine  is  constant, 
the  electromotive  force  of  the  system  decreases  as  the  load  increases. 

(c)  With  constant  rotor  speed,  the  frequency  of  the  system  de- 
creases as  the  load  increases. 

(d)  The  frequency  of  the  system  is  proportional  to  the  speed  of  the 
synchronous  machine. 

(e)  The  electromotive  force  of  the  system  is  increased  by  in- 
creasing the  field  excitation  of  the  synchronous  machine. 

(/)  The  load  on  the  system  is  proportional  to  the  speed  of  the 
rotor  above  synchronism,  i.e.,  above  the  speed  of  the  synchronous 
machine. 


224  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

21.  Parallel  operation  of  induction  generators.  —  The  relative 
speeds  of  two  or  more  synchronous  machines  connected  to  the  same 
system  must  remain  constant,  the  phase  relations  of  the  generated 
electromotive  forces  determining  the  ratio  of  load  division.*     The 
division  of  the  load  between  two  induction  generators  operating  in 
parallel  is  proportional  to  the  ratio  of  the  deviation  of  the  actual 
speeds  of  the  rotors  from  the  synchronous  speed. 

Since  the  speeds  of  two  induction  machines  operating  in  parallel 
need  not  have  a  constant  ratio  to  each  other,  induction  generators 
need  not  be  synchronized  but  simply  brought  up  to  approximately 
synchronous  speed,  and  the  switch  connecting  the  incoming  machine 
to  the  system  closed.  After  closing  the  switch,  the  load  division  is 
adjusted,  as  in  the  case  of  synchronous  machines,  by  manipulating 
the  speed  governing  apparatus  of  the  prime  mover. 

22.  Commercial   applications.  —  The   induction   generator   has 
not,  up  to  the  present  time,  come  into  any  extensive  use  f  and  must 
be  considered  as  still  in  the  stage  of  development.     Theoretically 
it  offers  decided  advantages  when  driven  by  water  wheels,  gas 
engines  or  other  prime  movers,  the  speed  regulations  of  which  are 

not  good. 

CHAPTER  XIII  —  PROBLEMS 

1.  The  maximum  torque  of  an  induction  motor  occurs  when  the  slip  is  18 
per  cent.     Find  the  ratio  of  the  rotor  resistance  and  the  resistance  that  must  be 
added  to  the  rotor  circuit  so  maximum  torque  is  developed  at  starting. 

Note.  —  The  starting  effort  of  an  induction  motor  is  expressed  in  "synchro- 
nous watts"  or  "synchronous  horse  power,"  and  is  equal  to  the  power  that 
would  be  developed  in  the  rotor  if  it  were  operating  at  synchronous  speed  and 
developing  a  torque  equal  to  that  developed  at  starting. 

2.  Find  the  frequency  of  the  rotor  current  in  a  lo-pole,  6o-cycle  induction 
motor  when  operated  at  a  speed  of:  (a)  720  r.p.m.,  (6)  600  r.p.m.,  (c)  400  r.p.m., 
(d)  100  r.p.m.,  (e)  o  r.p.m. 

3.  Compare  the  efficiencies  of  two  induction  motors,  the  full-load  slip  of  one 
being  4  per  cent  and  that  of  the  other  8  per  cent. 

4.  A  4oo-horse-power,  440- volt,  6o-cycle,  3-phase  induction  motor  was  tested 
and  the  following  data  obtained:          Stator  current       Watts  input 

per  phase  per  phase 

Without  load  160  4000 

With  rotor  blocked  2250  225000 

Stator  resistance  per  phase  =  0.018  ohm. 
Construct  the  circle  diagram. 
*  See  Chapter  9,  Section  12. 

t  The  59th  Street  power  house  of  the  Interborough  Rapid  Transit  Co.  (New  York) , 
has  an  installation  of  induction  generators.  So  far  as  the  writer  has  been  able  to  learn, 
this  installation  has  been  entirely  satisfactory. 


THE   INDUCTION  MOTOR  225 

5.  From  the  circle  diagram  of  Problem  4  plot  the  following  curves:  (a) 
speed-torque,  (6)  efficiency,  (c)  power  factor. 

6.  From  the  circle  diagram  of  Problem  4  determine  the  following  when  the 
output  of  the  motor  is  400  brake  horse  power:    (a)  current  input,  (b)  torque 
developed  in  the  rotor,  (c)  power  factor,  (d)  speed,  (e)  "added"  copper  losses 
in  stator,  (/)  rotor  losses. 

7.  From  the  circle  diagram  in  Problem  4  determine:    (a)  the  stator  current 
required  to  produce  maximum  starting  torque,  and  the  value  of  this  starting 
effort  in  synchronous  horse  power,  (b)  the  stator  current  and  the  torque  when 
the  power  factor  is  maximum. 

8.  Determine  the  synchronous  speed  of  a  6o-cycle  induction  motor  having: 
(a)  2  poles,  (b)  4  poles,  (c)  6  poles,  (d)  10  poles,  (e)  16  poles,  (/)  24  poles. 

9.  Determine  the  synchronous  speed  of  a  25-cycle  induction  motor  having 
poles  as  indicated  in  Problem  8. 

10.  Determine  the  voltages  induced  in  the  rotor  circuit  of  the  motor  in  Prob- 
lem 2  when  the  ratio  of  the  stator  turns  to  the  rotor  turns  is  one,  and  the  applied 
voltage  is  230. 

n.  A  6-  and  a  lo-pole  induction  motor  are  operated  in  cascade,  the  6-pole 
motor  being  connected  to  a  oo-cycle  alternating-current  circuit.  Find  the  syn- 
chronous speed  of  the  combination. 

Note.  —  When  two  induction  motors  are  to  be  operated  in  cascade  or  con- 
catenation, their  rotors  are  connected  to  the  same  shaft;  the  stator  of  one  motor 
is  supplied  directly  from  the  line,  and  the  stator  of  the  other  motor  from  the  rotor 
of  the  first  motor.  The  motors  may  be  connected  in  direct  or  in  differential 
cascade,  and  the  synchronous  speed  of  the  combination  is  that  of  a  single 
motor  having  the  sum  or  difference  of  the  number  of  poles  on  the  two  motors, 
and  operated  from  a  line  of  the  same  frequency  as  that  to  which  the  first  motor 
in  the  cascade  combination  is  connected. 

12.  Find  the  voltage  supplied  to  the  stator  windings  of  the  lo-pole  motor 
in  Problem  n. 

13.  Find  the  frequency  of  the  current  in  the  stator  windings  of  the  lo-pole 
motor  in  Problem  n. 

14.  A  6o-cycle  alternator  is  direct-connected  to  a  6-pole,  25-cycle  induction 
motor.     Find  the  number  of  poles  on  the  alternator. 

15.  An  induction  motor  develops  a  starting  torque  of  250  foot-pounds  when 
the  applied  voltage  is  60  per  cent  of  the  rated  voltage.     Determine  the  starting 
torque  that  would  be  developed  by  the  motor  if  rated  voltage  were  applied  to 
the  stator  windings. 

Note.  —  Since  the  fundamental  equation  of  the  induction  motor  is  the  same 
as  that  of  the  transformer  (E  =  4.44/4>Afio~8)  with  the  addition  of  a  winding 
constant,  both  the  flux  and  the  rotor  current  are  proportional  to  the  applied 
voltage.  Therefore,  the  torque  of  an  induction  motor  is  proportional  to  the 
square  of  the  applied  voltage,  the  slip  of  the  rotor  (the  frequency  of  the  rotor 
currents)  remaining  constant. 


CHAPTER  XIV 

SINGLE-PHASE   COMMUTATING   MOTORS 

i.  Action  of  the  shunt  motor.  —  When  alternating  current  is 
supplied  to  a  shunt  motor,  two  very  noticeable  effects  take  place: 
(a)  the  motor  develops  very  little  torque,  (b)  excessive  sparking 
occurs  at  the  brushes. 

(a)  Torque.  —  It  is  evident  that  if  the  current  in  the  armature  of 
a  shunt  motor  supplied  from  a  single-phase  alternating-current 
system  and  the  flux  set  up  by  the  field  winding  are  in  phase,  i.e., 
attain  their  maximum  and  minimum  at  the  same  instant,  a  torque 
varying  from  zero  to  maximum  but  always  in  the  same  direction 
is  produced.  The  average  torque  of  an  alternating-current  shunt 
motor  is  proportional  to  the  average  product  of  the  armature 
current  and  the  flux  in  the  air  gap,  as  in  the  continuous-current 
motor. 

When  the  field  winding  of  a  shunt  motor  is  excited  from  an  alter- 
nating-current system,  the  current  lags  behind  the  electromotive 
'e.m.f  force  by  a  very  considerable  angle,  the  field  circuit 
being  highly  inductive,  and  the  flux  has  a  correspond- 
ing lag  behind  the  electromotive  force.  The  power 
Fields  factor  of  the  armature  circuit  being  higher  than  that 
of  the  field  circuit,  the  vector  of  armature  current 

FIG.  183.       leads  the  vector  of  field  flux,  as  shown  in  Fig.  183. 

Because  of  the  phase  relations  of  the  armature  current  and  the 
field  flux,  their  product  is  negative  during  part  of  the  cycle  and 
positive  during  the  remainder  of  the  cycle.  The  torque,  therefore, 
tends  to  rotate  the  armature  first  in  one  direction  and  then  in 
the  other,  and  the  net  torque  producing,  or  tending  to  produce, 
rotation  is  the  algebraic  sum  of  the  average  positive  and  negative 
torques  during  one  complete  cycle.  If  the  angle  0  equals  90  degrees, 
the  sum  of  the  instantaneous  torques  is  zero,  and  the  armature  has 
no  tendency  to  rotate. 

The  torque  of  a  shunt  motor  may  be  improved  by  connecting  the 
armature  to  one  phase,  and  the  field  to  the  other  phase  of  a  two- 

226 


SINGLE-PHASE   COMMUTATING  MOTORS 


227 


FIG.  184. 


phase  system.  Such  a  machine  is  seldom  used  because  of  the  com- 
plications involved,  and  the  low  power  factor  at  which  the  phase 
supplying  the  field  circuit  must  operate,  and  the  fact  that  more  satis- 
factory apparatus  has  been  devised. 

(b)  Sparking.  —  Commutation  in  an  alternating-current  motor  is 
more  complicated  than  in  a  continuous-current  dynamo.  Let  the 
position  of  the  armature  of  an  alternating-current  motor  be  that 
shown  in  Fig.  184,  the  armature  coil  c  being  short-circuited  by  the 
brush.  The  changing  value  of  flux  in  the  armature  core  causes 
the  coil  c  to  act  as  the  short-circuited  second- 
ary of  a  transformer  in  which  the  current  may 
be  many  times  the  normal  current  flowing 
in  the  coil.  This  large  current  flowing  in  the 
local  circuit,  composed  of  the  coil,  the  two 
commutator  segments  to  which  the  coil  is  con- 
nected, and  the  brush,  causes  excessive  heating  of  the  armature  as 
well  as  destructive  sparking  when  the  brush  passes  from  one  com- 
mutator segment  to  another. 

2.   The  series  motor.  —  When  a  continuous-current  series  motor 
is  operated  with  alternating  current,  the  phase  displacement  between 

the  armature  current  and  the  field 
flux  largely  disappears,  since  the 
same  current  flows  in  both  wind- 
ings and  the  inductance  of  a  series 
field  winding  is  much  less  than 
that  of  a  shunt  field  winding,  but 
commutation  difficulties  still  exist. 
The  commutation  of  a  series 
motor  operating  with  alternating 
current  is  improved  by:  (a)  re- 
ducing the  flux  density  of  the 
magnetic  circuit,  (b)  reducing  the  number  of  series  turns  per  arma- 
ture coil,  (c)  reducing  the  frequency  of  the  supply  circuit,  (c)  special 
devices. 

(a)  Flux  density.  —  For  a  given  frequency,  the  average  rate  at 

which  the  flux  changes  is  proportional  to  its  maximum  value,  and 

the  electromotive  force  induced  in  the  short-circuited  armature  coil 

is  proportional  to  the  rate  of  change.*    The  impedance  of  the 

*  See  Chapter  n,  Section  i. 


FIG.  185.    The  Series  Motor. 


228  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

short-circuit  remaining  constant,  the  current  in  the  local  circuit 
is  proportional  to  the  induced  electromotive  force. 

(b)  Series  turns.  —  Since  the  electromotive  force  induced  in  the 
secondary  of  a  transformer  is  proportional  to  the  number  of  series 
turns,*  the  smaller  the  number  of  series  turns  per  armature  coil, 
the  less  is  the  induced  voltage.     It  might   appear  that   reducing 
the  number  of  turns  in   the  coil   reduces  the  impedance  of  the 
circuit  in  the  same  ratio,  and  that  the  current,  therefore,  remains 
constant;  but  the  resistance  of  the  brush,  brush  contact,  commu- 
tator bar,  and  connecting   leads  must  be  taken  into  account  in 
calculating  the  impedance  of  the  local  circuit. 

(c)  Frequency.  —  The  rate  of  flux  change  is  proportional  to  the 
frequency,  i.e.,  the  higher  the  frequency  the  shorter  the  time  in 
which  the  flux  must  change  from  zero  to  maximum.     The  electro- 
motive force  induced  in  a  short-circuited  coil  is,  therefore,  pro- 
portional to  the  frequency  of  the  supply  system,  and  low  frequencies 
tend  to  reduce  commutation  troubles.     Series  alternating-current 
motors  are  seldom  used  on  circuits  having  a  frequency  greater  than 
twenty-five  cycles  per  second. 

(d)  Special  devices.  —  While  the  above  considerations  of  design 
and  operation  materially  reduce  both  heating  and  commutation 
troubles,  special  devices  have  been  found  necessary  to  bring  them 
within  practical  operating  limits.     The  simpler  (and  more  satis- 
factory) of  these  devices  are:  (i)  resistance  leads  connected  be- 

winding  tween    the    terminals    of    the    armature 

coils  and  the  commutator  segments,  (2) 
balanced  choke  coils. 

(i)  Resistance    leads.  —  If    resistances 

Comma fa/or  .  . 

are  connected  as  indicated  in  Fig.  i86a, 
FIG.  i86a.    Resistance  Leads   the  local  current  is  reduced  correspond- 

A.  C.  Series  Motor.  ingly  since  jt  must  flow  through  the  coil 

and  two  resistances  connected  in  series.  The  load  current  must, 
also,  flow  through  these  resistances  and  this  would,  apparently, 
decrease  the  efficiency  by  increasing  the  resistance  losses.  It  is  an 
experimental  fact  that  the  introduction  of  resistance  leads  increases 
the  efficiency,  the  increased  loss  due  to  the  load  current  flowing 
in  the  added  resistance  being  less  than  the  decreased  loss  due  to  the 
smaller  current  flowing  in  the  local  circuit. 

*  See  Chapter  n,  Section  3. 


SINGLE-PHASE  COMMUTATING  MOTORS 


22Q 


Armature  Winding 


Brush 


FIG.  i86b.     Balanced  Choke 
Coils. 


Winding 


(2)  Choke  coils.  —  The  connections  for  the  use  of  choke  coils  are 
shown  in  Fig.  i86b.  The  windings  on  each  core  are  so  related 
that  their  inductance  is  cumulative  to 
the  short-circuit  current,  but  differen- 
tial to  the  load  current.  This  arrange- 
ment is  not  entirely  satisfactory  because 
the  coils  balance  for  the  load  current 
only  when  the  current  is  equally  divided 
between  two  coils,  and  for  certain 
positions  of  the  armature,  the  coils 
neutralize  each  other,  and  become  ineffective  so  far  as  the  short- 
circuit  current  is  concerned. 

3.  Compensation  for  armature  inductance.  —  The  inductance 
of  the  armature  winding  may  be  neutralized  by  means  of  a  "  com- 
pensating" winding  connected  as  shown  in  Fig.  187.     The  com- 
pensating  winding    may   be    sup- 
plied with  current :  (a)  inductively, 
(b)  conductively. 

(a)  Current  supplied  inductively. 
-  If  the  compensating  winding 
is  closed  on  itself  (short-circuited) 
as  indicated  in  Fig.  i8ya,  the 
alternating  field  flux  induces  in 
the  winding  an  electromotive  force, 
and  the  magnetic  effect  of  the 
winding  is  approximately  equal  and 
opposite  to  that  of  the  armature 
winding.  The  magnetic  fields  of 
the  two  windings,  therefore,  neu- 
tralize each  other. 
(b)  Current  supplied  conductively.  —  If  the  compensating  winding 
is  connected  in  series  with  the  armature  winding  as  indicated  in 
Fig.  i8yb,  the  same  current  flows  in  the  windings.  By  properly 
proportioning  the  compensating  winding,  and  connecting  it  so  that 
its  magnetic  effect  is  opposite  to  that  of  the  armature  winding,  the 
magnetic  fields  of  the  two  windings  neutralize  each  other. 

4.  Comparison  of  series  motors.  —  The  alternating-current  and 
the  continuous-current  series  motors  differ  in  the  following  re- 
spects : 


(b)  Conductive  Compensation 

FIG.  187. 


Field  Winding 


230 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


rmafure  Drop. 
Drvp  in  Compensating 
Winding 


FIG.  1 88.     Clock  Diagram,  Series 
A.  C.  Motor. 


(a)  The  weight  of  the  alternating-current  motor  is  from  one  and 
one-half  to  two  times  that  of  the  continuous-current  motor  develop- 
ing the  same  torque. 

(b)  The  alternating-current  motor  has 
the  larger  number  of  armature  coils. 

(c)  The  alternating-current  motor  has 
the  larger  commutator,  and  the  larger 
number  of  segments. 

(d)  The  entire  magnetic  circuit  of  the 
alternating-current  motor  is  laminated. 

5.  The  repulsion  motor.  —  When  a  continuous-current  armature 
is  placed  in  a  magnetic  field  produced  by  an  alternating  current  as 
indicated  in  Fig.  189,  the  field  wind- 
ing acts  as  the  primary  and  the 
armature  winding  as  the  second- 
ary of  a  transformer,  and  current 
flows  between  the  short-circuited 
brushes  for  any  position  of  the 
brushes  except  that  shown  in  Fig. 
i8gb,  when  the  algebraic  sum  of 
the  voltages  induced  in  the  coils 
between  brushes  is  zero. 


{  FIG.  189.    Repulsion  Motor. 

For  the  position  indicated  in  Fig.  i8ga,  maximum  current  flows 
between  the  brushes.  This  current  tends  to  set  up  a  flux  dia- 
metrically opposed  to  that  set  up  by  the  field  windings,  and  the 
magnetic  effect  of  the  field  windings  is  largely  neutralized.  The 
torque  produced  by  the  reaction  between  any  current-carrying 
conductor  and  the  field  flux  is  neutralized  by  an  equal  and  oppo- 


SINGLE-PHASE   COMMUTATING  MOTORS 


23I 


site  torque  produced  by  some  other  armature  conductor  and  the 
same  field  flux.  The  armature  has,  therefore,  no  tendency  to 
rotate. 

This  equality  of  opposite  torques  is  destroyed,  and  the  armature 
caused  to  rotate,  by  moving  the  brushes  in  either  direction,  the 
direction  of  rotation  being  the  same  as  the  movement  of  the 
brushes.  It  has  been  determined,  experimentally,  that  maximum 
torque  is  developed  when  the  angle  0  equals  approximately  45 
degrees. 

While  the  repulsion  motor  is  little  used  commercially,*  its  good 
starting  torque  offers  means  by  which  the  single-phase  induction 
motor  may  be  made  self-starting.  The  Wagner  Electric  Mfg.  Co. 
make  a  motor  of  this  type  in  which  the  rotor  structure,  at  starting, 
is  essentially  that  described  above.  When  the  speed  reaches  a 
predetermined  value,  a  centrifugal  device  removes  the  brushes  from 
the  commutator,  and  short-circuits  the  commutator  bars  so  that 
the  armature  conductors  form  a  squirrel-cage  structure. 

6.  The  compensated  repulsion 
motor.  -  -  The  so-called  compen- 
sated repulsion  motor  is,  mechan- 
ically, a  series  motor  with  the 
addition  of  short-circuited  brushes 
placed  in  quadrature  with  the  main 
brushes,  as  indicated  in  Fig.  190. 
While  both  series  and  repulsion 
motors  have  series  characteristics, 
i.e.,  the  speed  tends  to  rise  to  an 
infinite  value  as  the  load  approaches 


FIG.  190. 


Compensated  Repulsion 
Motor. 


zero,  the  compensated  repulsion  motor  has  shunt  characteristics, 
and  its  principles  of  operation  are  materially  different  from  either 
the  series  or  the  repulsion  motor. 

The  short-circuited  brushes  of  this  type  of  motor  cause  the  arma- 
ture winding  to  produce  a  magnetic  field  which  opposes  and  largely 
neutralizes  the  flux  set  up  by  the  field  winding.  The  current 
which  flows  between  the  short-circuited  brushes  is  produced  by 
transformer  action,  as  described  for  "the  repulsion  motor  when  the 
brushes  are  in  the  position  indicated  in  Fig.  iSpa. 

The  current  flowing  between  the  main  (series)  brushes  sets  up  a 

*  Commutation  is  inherently  poor. 


ESSENTIALS   OF  ELECTRICAL   ENGINEERING 


field  at  right  angles  to  the  line  joining  the  short-circuited  brushes. 
This  flux,  reacting  with  the  current  flowing  in  the  armature  con- 
ductors by  reason  of  the  short-circuited  brush  connection,  produces 
the  greater  part  of  the  torque  of  the  motor,  although  some  torque 
is  doubtless  produced  by  a  reaction  between  the  series-field  flux 
and  the  current  between  the  main  (series)  brushes. 

As  the  armature  speed  increases,  the  counter-electromotive  force 
induced  in  the  conductors  by  reason  of  their  motion  reduces  the 
current  flowing  between  the  short-circuited  brushes  and  the  torque 

becomes  less.  Therefore,  the  flux  is 
approximately  constant,  the  current 
varies  inversely  with  the  speed  of  the 
armature  and  the  motor  has  shunt 
characteristics. 

7.   The   General  Electric   RI   motor. 
-  The    connections     of     the     General 
FIG.  191.    Diagram  of  General    Electric    Company's    Type    RI    motor 
Electric  "RI"  Motor.  •     £•  •     i 

are  shown  in  Fig.  191,  and  typical  per- 
formance curves  in  Fig.   192.     From  the  curve  sheet,  it  will  be 


Typical  Performance  Curves 
Type  RI  -  4  Pole  1600  ff.  P.  M. 
Repulsion  Induction  Motor 


73  100 

Percent  Load 

FIG.  192.     Typical  Performance  Curves  of  Type  RI  Motor 

observed  that  the  power  factor  is  high  for  loads  above  50  per  cent 
of  the  rated  output. 

8.  The  Wagner  BK  motor.  —  The  Unity  Power  Factor  (so- 
called  by  the  manufacturers)  single-phase  motor  of  the  Wagner 
Electric  Mfg.  Co.  is  a  combination  of  the  compensated  repulsion 


SINGLE-PHASE   COMMUTATING  MOTORS 


233 


motor  and  the  single-phase  induction  motor.  The  armature  has 
a  squirrel-cage,  as  well  as  a  commutated  winding,  as  indicated  in 
Fig.  193.  The  electrical  connections  are  indicated  in  Fig.  194. 


Retaini 

Commuted  Winding  (3) 

Magnetic  Separator 

Squirrel  Ca^e  (4) 


FIG.  193.     Section  Through  Slot  of  Wagner 
"BK"  Motor. 


FIG.  194.    Wiring  Diagram  for  Wag- 
ner "  BK  "  Motor. 


At  starting,  the  main  field  winding  i  induces,  by  transformer 
action,  currents  in  the  commutated  winding  3,  and  these  currents 
flow  between  the  short-circuited  brushes  5  and  6.  Similarly,  cur- 
rents are  induced  in  the  squirrel-cage  winding  as  described  for  the 
single-phase  induction  motor.*  The  series  currents  flowing  in 


14  V  BK  MOTOR 
H  P.60~.1  PHASE,  4  POLE 
220  VOLTS.  18OO  R.P.M. 


•»"'  123456789       10  H.P. 

FIG.  195.    Performance  Curves  of  Wagner  "BK"  Motor. 

the  commutated  winding  3  set  up  a  flux,  the  direction  of  which  is 
at  right  angles  to  the  flux  set  up  by  winding  i.  The  currents  in 
the  commutated  and  in  the  squirrel-cage  windings  react  with  this 
quadrature  flux  to  produce  a  torque,  and  start  the  motor.  As  the 
speed  of  the  motor  increases,  the  squirrel-cage  winding  sets  up  a 
quadrature  flux  of  its  own,  and  develops  a  corresponding  torque. 

*  See  Chapter  13,  Section  14. 


234  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

The  torque  of  this  motor  is,  then,  a  resultant  of  two  torques,  one 
of  which  is  maximum  at  starting  and  decreases  as  the  speed  in- 
creases, while  the  other  is  zero  at  starting,  increases  to  maximum  as 
the  speed  increases,  and  then  decreases  as  the  speed  increases 
still  further. 

The  action  of  the  auxiliary  winding  2,  is  to  improve  the  power 
factor  of  the  motor.  The  switch  9  is  open  at  starting,  but  is  auto- 
matically closed  by  a  centrifugal  device  at  a  predetermined  speed. 

The  performance  curves  of  this  motor  are  shown  in  Fig.  1 95 .  The 
power  factor  will  be  observed  to  be  70  per  cent  leading  at  zero  load, 
and  approximately  unity  from  75  to  150  per  cent  of  rated  load.  It 
will  also  be  noted  that  the  speed  is  very  nearly  synchronous  at  rated 
load,  rises  slightly  above  synchronism  at  no-load,  and  drops  below 
synchronism  as  the  load  is  increased  above  the  rated  capacity. 


CHAPTER  XV 
ELECTRIC   LAMPS* 

i.  Arc  lamps. —  Arc  lamps  are  largely  used  for  lighting  streets 
and  other  areas  where  a  few  large  units  are  preferable  to  a  larger 
number  of  smaller  units. 

The  basic  principle  of  the  arc  lamp  is  the  arc  which  is  established 
when  an  electric  circuit  is  interrupted.  The  heat  of  the  arc  causes 
one  or  both  electrodes  to  be  consumed,  and  the  incandescent  par- 
ticles emit  a  very  intense  light.  Arc  lamps  may  be  classified  as :  (a) 
carbon  arcs,  (b)  magnetite  arcs,  (c)  flaming  arcs.  Carbon  and 
flaming  arc  lamps  may  be  operated  on  either  alternating  or  unidi- 
rectional current;  magnetite  lamps  on  unidirectional  current  only. 

Arc  lamps  may  be  operated  either  in  parallel  or  in  series.  In 
parallel  operation,  the  voltage  is  constant  and  the  regulating  mech- 
anism must  maintain  the  proper  value  of  current;  in  series  oper- 
ation, the  current  is  automatically  maintained  at  a  constant  value, 
and  the  lamp  mechanism  controls  the  voltage  between  the  terminals 
of  the  lamp. 

(a)  Carbon  arc  lamps.  —  In  this  type  of  lamp  the  electrodes  are 
made  of  finely  ground  carbon  mixed  with  a  suitable  binder,  pressed 
into  the  form  of  pencils  and  baked.  In  the  early  lamps,  the  arcs 
were  "open,"  i.e.,  air  currents  circulated  freely  about  the  arc.  The 
light  from  such  a  lamp  is  unsteady  and  the  electrodes  are  consumed 
very  rapidly.  By  enclosing  the  arc  in  a  glass  globe  to  which  a  very 
limited  quantity  of  air  is  admitted,  the  flickering  of  the  arc  is  materi- 
ally reduced,  and  the  electrodes  are  consumed  at  a  much  slower  rate. 
Because  of  these  facts  and  the  reduced  fire  hazard  of  the  enclosed 
arc,  open  arc  lamps  are  no  longer  manufactured. 

Fig.  196  shows  the  circuits  of  a  lamp  designed  for  parallel  opera- 
tion. Regulation  is  accomplished  by  means  of  an  electromagnet, 

*  Problems  in  illumination  are  beyond  the  scope  of  this  volume.  For  information 
regarding  methods  of  calculating  light  densities,  etc.,  the  reader  is  referred  to  "  Illum- 
ination and  Photometry,"  by  W.  E.  Wickenden,  and  "  Electrical  Illuminating  Engineer- 
ing," by  W.  E.  Barrows,  Jr. 

235 


236 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


Line 

Resistance  \ 

: 

:  Series  Coil 

r 

Carbon 
•Clutch 

*T 

T  'heat  Sere 
Carbon 

•en 

the  windings  of  which  are  connected  in  series  with  the  arc.  When 
the  circuit  is  open,  the  core  of  the  magnet  drops  to  its  lowest  position 
and  the  clutch  releases  the  upper  electrode,  which  then  rests  on  the 

lower  electrode  as  indicated.  Closing  the 
lamp  circuit,  causes  the  magnet  to  attract 
its  core,  raise  the  upper  electrode,  and 
." strike"  the  arc. 

The  resistance  of  the  circuit  becomes 
greater  as  the  distance  between  the  elec- 
trodes increases,  and  the  magnet  is   so 
proportioned  that  its  pull  balances  that 
FIG.  196.    Schematic  Diagram  of  a  spring,  when  rated  current  flows  in 

of   Connections  for  Parallel    ^  ^^       The  ^     therefo        moves 
(Carbon)  Arc  Lamp. 

upward    until    this    equilibrium,    which 

corresponds  to  a  fixed  distance  between  the  electrodes,  is  estab- 
lished. As  the  electrodes  burn  away,  the  length  of  the  arc 
increases,  the  current  in  the  circuit  decreases,  the  pull  of  the 
magnet  no  longer  balances  that  of  the  spring,  and  the  core  moves 
downward,  increasing  the  current  and  decreasing  the  length  of 
the  arc,  until  equilibrium  is  restored.  When  the  core  has  de- 
scended to  its  lower  limit,  the  clutch  releases  the  upper  electrode 
which  drops  into  contact  with  the  lower  electrode,  the  resistance  of 
the  circuit  is  reduced,  and  the  increased  

Line 

current  causes  the  magnet  to  separate 
the  electrodes  again.  The  result  of  this 
process  is  the  periodical  feeding  of  the 
upper  electrode  through  the  clutch  by 
an  amount  equal,  approximately,  to 
the  length  of  the  arc. 

Since  multiple  arc  lamps  are  usually 
connected  to  no-volt  mains,  and  the  _ 

FIG.  197.     Schematic  Diagram  of 

arc  requires  only  from  seventy  to  eighty      connections  for  Series  (Carbon) 
volts,  the  lamp  is  provided  with  a  ballast      Arc  Lamp.    (Differential  Con- 
by  means  of  which  the  line  voltage  is  re- 
duced.    In  continuous- current  lamps  the  ballast  is  resistance;   in 
alternating-current  lamps,  reactance. 

The  series  lamp  (Fig.  197)  has,  in  addition  to  the  series  magnet 
of  the  parallel  lamp,  a  magnet  connected  in  parallel  with  the  arc  and 
so  arranged  that  its  action  opposes  that  of  the  series  magnet.  When 


Heat  Screen 
.Carbon 


ELECTRIC  LAMPS 


237 


current  flows  through  the  lamp,  the  series  magnet  raises  the  upper 
electrode  as  in  the  parallel  lamp,  but  as  the  length  of  the  arc  in- 
creases, the  current  in  the  windings  of  the  shunt  magnet  increases, 
and  equilibrium  is  established  at  the  length  of  arc  at  which  the  pull 
of  the  series  magnet  (constant)  minus  the  pull  of  the  shunt  magnet, 
balances  the  pull  of  a  spring.  As  the  electrodes  burn  away,  the 
effect  of  the  shunt  magnet  increases,  and  equilibrium  is  main- 
tained by  the  changing  position  of  the  series  magnet  core.  At  the 
lower  limit  of  the  movement  of  the  core,  the  clutch  releases  the 
upper  electrode  and  it  is  fed  downward  as  in  the  parallel  lamp. 

In  the  series  lamp,  a  bypass  must  be  provided  for  the  passage  of 
the  current  in  case  the  carbons  burn  out,  or  for  any  other  reason  the 
arc  circuit  becomes  opened.  This  may  be  provided  by  means  of  a 
shunt  coil  armature  which  closes  the  circuit  through  an  auxiliary 
resistance  whenever  the  voltage  across  the  arc  reaches  a  given  value, 
which  is  always  higher  than  that  reached  in  normal  operation. 
Should  the  arc  circuit  be  re-established,  the  voltage  between  the 
terminals  of  the  shunt  winding  is  reduced,  the  armature  of  the 
shunt  magnet  is  released  and  the  auxiliary  circuit  broken. 

(b)  Magnetite  (or  "luminous")  arc  lamps.  —  The  magnetite  arc 
lamp  is  one  of  the  results  of  the  researches  of  Dr.  C.  P.  Steinmetz  of 
the  General  Electric  Co.,  and  has  a  fixed  upper  electrode  *  of  copper 
and  a  movable  lower  electrode  composed  of  . 

Line 

a  mixture  of  magnetite  (one  of  the  oxides 
of  iron)  and  titanium  oxide  encased  in  an 
iron  tube.  This  lamp  is  economical  in 
power  consumption  and  gives  a  brilliant 
" white"  light  which  is  well  distributed. 

Fig.  198  shows  the  circuits  of  a  series 
magnetite  lamp,  and  its  operating  mechan- 
ism.  Unlike  the  carbon  lamp,  the  arc  circuit  FlG  Ig8.  Schematic  Diagram 
is  open  when  the  lamp  is  not  in  operation.!  of  Connections  for  Series 
When  the  current  is  turned  on,  the  starting 
magnet  closes  the  arc  circuit  by  raising  the 
lower  electrode  into  contact  with  the  copper  rod.  The  series  mag- 
net then  attracts  its  armature  and  opens  the  circuit  of  the  starting 


Snitch 


Resistance 


'  Copper  Electrode 
'Magnetite  Electrode 


Magnetite  (Luminous)  Arc 
Lamp. 


*  In  the  Westinghouse  magnetite  lamp  the  lower  electrode  is  fixed, 
t  If  the  arc  circuit  is  not  kept  open  the  terminals  weld  or  stick  together,  making 
starting  difficult  or  impossible. 


238  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

magnet,  allowing  the  lower  electrode  to  drop  and  "strike"  the  arc. 
As  the  electrode  burns  away,  the  voltage  over  the  arc  increases  until 
the  shunt  magnet  attracts  its  armature,  closes  the  circuit  of  the 
starting  magnet,  and  feeds  the  lower  electrode  upward  by  a  definite 
step.  In  case  the  arc  circuit  becomes  open,  the  shunt  magnet  closes 
the  circuit  of  the  starting  magnet  and  the  current  flows  through  the 
starting  magnet  and  its  series  resistance,  so  that  the  other  lamps  in 


(a)  Complete.  .  (b)  Casing  Removed. 

FIG.  199.     G.  E.  Series  Luminous  (Magnetite)  Arc  Lamp. 

the  circuit  are  unaffected  by  the  burning  out  of  an  electrode,  or  by 
failure  to  feed  properly. 

The  mechanism  of  the  magnetite  lamp,  when  intended  for  parallel 
operation,  is  similar  to  that  of  the  series  lamp,  with  the  omission  of 
the  shunt  magnet.  The  starting  magnet  raises  the  lower  electrode 
and  the  series  magnet  breaks  the  circuit  of  the  starting  magnet, 
which  allows  the  lower  electrode  to  fall  and  strike  the  arc  as  in  the 
series  lamp.  When  the  electrode  has  burned  away  to  such  an  ex- 
tent that  the  series  coil  is  no  longer  able  to  hold  its  armature,  the 
circuit  of  the  starting  magnet  is  closed,  and  the  electrode  fed  upward. 

(c)  Flaming-arc  lamps.  —  The  light  of  the  flaming-arc  lamp  is 
due  to  the  luminescent  vapor  of  metallic  salts,  the  salts  of  calcium 
(yellow  light)  and  of  titanium  (white  light)  being  those  com- 
monly used.  The  electrodes  consist,  essentially,  of  a  mixture  of 
carbon,  the  metallic  salt,  and  an  alkaline  salt.  The  purpose  of  the 
alkaline  salt  is  to  steady  the  arc  and  to  prevent  the  accumulation 
of  slag. 


ELECTRIC  LAMPS  239 

High  efficiency  is  obtained  only  by  rapid  consumption  of  the 
electrodes,  the  great  length  (14  to  24  inches)  and  small  diameter  (J 
to  f  inch)  of  which  make  vertical  mounting  impracticable.  The 
electrodes  are,  therefore,  usually  placed  as  indicated  in  Fig.  200,  and 
fed  downward  by  gravity,  the  mechanism  holding  them  being  re- 
leased by  an  electromagnet  connected  in  parallel  with  the  arc,  when 
the  voltage  across  the  arc  reaches  a  predetermined  value.  The 
electrical  circuits  of  one  of  the  simpler  regulating  mechanisms  are 
shown  in  Fig.  200,  and  are  similar  to  those  of  the  differential  carbon 
lamp  described  above.  The  electrodes  are  normally  separated. 
When  the  circuit  is  closed,  the  shunt  magnet  pulls 
the  points  of  the  electrodes  together  (lateral  move- 
ment of  one  electrode  is  provided  for),  closes  the 
arc  circuit,  and  energizes  the  series  magnet.  The 
series  magnet  separates  the  electrodes,  thus  striking 
the  arc,  and  equilibrium  is  established  when  the  arc 
is  from  ij  to  2  inches  in  length.  As  the  electrodes  FIG.  200.  Wiring 
burn  away,  the  voltage  over  the  shunt  coil  increases,  Diagram  Flam- 
while  the  current  in  the  series  coil  tends  to  decrease  mg  Arc  Lamp' 
(parallel  operation),  and  equilibrium  is  maintained  by  an  inward 
movement  of  the  movable  electrode.  When  the  electrode  reaches 
the  limit  of  its  movement,  the  shunt  magnet  causes  the  mechanism 
holding  the  electrodes  to  be  released,  and  they  are  fed  downward, 
reducing  the  length  of  the  arc.  The  increased  current  in  the  wind- 
ings of  the  series  magnet  causes  the  electrodes  to  be  separated  and 
the  arc  is  restored  to  its  normal  length. 

The  short  life  (10  to  18  hours)  of  flaming-arc  electrodes  led  to  the 
development  of  the  regenerative  flaming-arc  lamp,  which  is  an 
adaptation  of  the  flaming-arc  principle  to  the  enclosed  lamp,  and 
increases  the  life  of  the  electrodes  to  seventy  hours  or  more.  Tubes 
through  which  the  gases  circulate  conserve  the  heat,  produce  a  more 
perfect  combustion,  and  increase  the  efficiency  of  the  lamp.  The 
electrodes  used  in  regenerative  lamps,  being  shorter  and  of  greater 
diameter  than  those  used  in  the  flaming-arc  lamps,  may  be  arranged 
vertically  as  in  the  carbon  and  magnetite  lamps. 

The  flaming-arc  is  the  most  efficient  of  artificial  illuminants,  and 
gives  approximately  three  candle  power  per  watt. 

Arc  lamps  are  always  provided  with  air  dash  pots  or  some  similar 
device  for  damping  the  movements  of  the  electrode,  which  would 
otherwise  be  jerky  and  irregular. 


240  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

2.  Instability  of  the  electric  arc.  —  The  electric  arc  is  inherently 
unstable  when  operated  from  constant  potential  mains,  i.e.,  when 
the  arc  is  connected  directly  between  the  mains.  Suppose  an 
electric  arc  to  be  established  between  constant  voltage  mains.  Any 
slight  decrease  in  the  value  of  the  current  flowing  in  the  circuit 
causes  the  cross-sectional  area  of  the  arc  to  decrease,  increases  the 
resistance  of  the  circuit,  and  still  further  decreases  the  current; 
any  slight  increase  in  the  value  of  the  current  flowing  in  the  circuit 
causes  the  cross-sectional  area  of  the  arc  to  increase,  decreases  the 
resistance  of  the  circuit,  and  still  further  increases  the  current.  The 
effects  of  any  momentary  change  in  the  value  of  the  current  flowing 
in  the  circuit  are,  therefore,  cumulative,  and  either  cause  the  arc  to 
" break,"  or  the  current  to  increase  to  an  excessive  value. 

Stable  operation  of  the  multiple  arc  lamp  is  established  by  means 
of  the  ballast  (resistance  in  the  continuous- current  lamp  and  react- 
ance in  the  alternating-current  lamp)  referred  to  above,  the  opera- 
tion of  which  is  as  follows:  If  the  current  decreases  slightly,  the 
drop  in  the  ballast  decreases,  and  the  voltage  between  the  terminals 
of  the  arc  increases;  if  the  current  increases  slightly,  the  drop  in  the 
ballast  increases,  and  the  voltage  between  the  terminals  of  the  arc 
decreases.  The  ballast  thus  produces  a  compensating  change  in 
the  voltage  between  the  terminals  of  the  arc  for  any  change  in  the 
resistance  of  the  circuit  (area  of  the  arc). 

3.  Power  factor  of  the  alternating-current 
arc.  —  It  is  an  experimental  fact  that  the  power 
factor  of  the  alternating-current  arc  is  less  than 
unity,  its  average  being  about  0.85.     This  low 
FIG.    201.      Distorted  P°wer  factor  is  not  due  to  a  time  lag  between 

Current  Wave  of  the  the  current  and  the  electromotive  force,  but  to 

Electric  Arc.  distortion  of  the  current  wave.     As  pointed  out 

in  Section  2,  the  resistance  of  the  electric  arc  is  a  function  of  the 
current  flowing  in  the  circuit,  and  the  resistance  of  an  alternating- 
current  arc  depends  on  the  instantaneous  value  of  the  current.  Fig. 
20 1  is  a  reproduction  of  oscillograms  of  the  current  and  electromotive 
force  waves  in  an  alternating-current  arc. 

4.  Incandescent  lamps.  —  The  incandescent  lamp,  which  is 
universally  used  for  interior  lighting,  consists  of  a  hair-like  filament 
enclosed  in  a  highly  exhausted  and  hermetically  sealed  glass  globe. 
The  light  of  the  lamp  is  an  indirect  effect,  and  is  due  to  the  fact  that 


ELECTRIC  LAMPS  241 

the  electric  current  raises  the  temperature  of  the  filament  to  a 
"white"  heat.  The  purpose  of  the  globe,  in  addition  to  the  me- 
chanical protection  which  it  gives,  is  to  retard  oxidation  of  the  fila- 
ment and  thus  increase  the  life  of  the  lamp.  Connection  between 
the  filament  and  the  line  conductors  is  made  through  short  wires 
embedded  in  the  glass. 

For  more  than  twenty-five  years  after  the  incandescent  lamp 
became  a  commercial  article,  carbon  was  practically  the  only  sub- 
stance of  which  filaments  were  made,  but  within  a  few  years  carbon 
filaments  have  been  largely  displaced  by  those  of  metallic  tungsten. 
The  light  produced  by  the  tungsten  lamp  is  decidedly  superior  to 
that  of  the  carbon  lamp,  and  approximately  three  times  as  much 
light  is  available  from  a  given  expenditure  of  energy.  The  per- 
fection of  the  tungsten  lamp,  which  is  sold  under  the  trade  name 
"  Mazda,"  has  greatly  reduced  the  cost  of  electric  light,  and  thereby 
largely  increased  its  use. 

While  the  efficiency  of  an  incandescent  lamp  increases  as  the 
applied  voltage  is  increased,  the  life  of  the  filament  is  shortened. 
Lamps  are  rated  at  that  voltage  which  has  been  found  to  give  the 
most  satisfactory  results,  taking  into  account  both  efficiency  and 
length  of  life,  and  should  be  operated  at  this  voltage. 

The  temperature,  and  therefore  the  efficiency,  at  which  a  tung- 
sten filament  may  be  operated  is  materially  increased  when  it  is 
surrounded  by  an  atmosphere  of  nitrogen.  The  nitrogen-filled 
tungsten  lamp  is  now  a  commercial  article  and  has  an  efficiency,  in 
large  units,  as  low  as  |  watt  per  candle  power. 

The  incandescent  lamp  is  a  constant  potential  lamp,  and  is  used, 
primarily,  for  parallel  connection  between  constant  voltage  mains, 
although  series  lamps  are  used  to  a  considerable  extent  in  street 
lighting. 

5.  The  Nernst  lamp.  —  The  luminous  element  (glower)  of  the 
Nernst  lamp  is  a  hollow  cylinder  composed  of  oxides  of  some  of 
the  rarer  elements  (zirconia,  yttria,  etc.).  The  glower  is  a  non- 
conductor at  ordinary  temperatures,  and  is  heated  to  a  conducting 
temperature  by  a  coil  of  platinum  wire  connected  in  parallel  with 
the  glower.  The  circuit  of  the  heating  coil  is  automatically  opened 
when  current  begins  to  flow  in  the  glower  circuit. 

The  negative  temperature  coefficient  of  the  glower  makes  the 
operation  of  the  lamp  unstable  unless  a  ballast  is  connected  in  series 


242  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

with  the  glower.     The  ballast  coil  is  made  of  iron,  the  temperature 
coefficient  of  which  is  positive. 

The  glowers, of  Nernst  lamps  are  always  enclosed,  usually  by 
alabaster  globes  which  conserve  the  heat  and  prolong  the  life  of  the 
glowers.  The  color  effects  of  this  lamp  are  good  and  the  efficiency 
is  high. 

6.  The  mercury  vapor  lamp.  —  The  mercury  vapor  lamp  con- 
sists of  a  highly  exhausted  glass  tube  containing  a  small  quantity 
of  mercury,  which  may  be  vaporized  by  an  electric  current.     The 
light,  which  is  due  to  the  luminescence  of  the  mercury  vapor,  is 
entirely  devoid  of   red  rays.     This    absence   of   red  rays   causes 
colors  to  be  distorted,  but  the  light  is  very  acceptable  in  draughting 
rooms,  packing   rooms,   warehouses   and  other   places  where   its 
peculiar  color  effects  are  not  objectionable.     There  is  now  on  the 
market  a  phosphorescent  reflector  which  adds  red  rays  to  the  light 
emitted  by  the  mercury  lamp. 

Like  the  mercury  rectifier,  to  which  it  is  analogous,  the  mercury 
vapor  lamp,  because  of  its  high  resistance  when  cold,  must  be  started 
by  an  auxiliary  device  of  some  kind.  Starting  may  be  effected 
by  tilting  the  tube  until  the  terminals  are  connected  by  liquid  mer- 
cury through  which  the  current  flows,  heating  the  mercury  and  fill- 
ing the  tube  with  mercury  vapor.  The  tube  is  tilted  by  hand,  or 
by  means  of  an  electromagnet  which  is  automatic  in  its  action. 
Starting  may  also  be  accomplished  by  breaking  an  auxiliary  circuit, 
the  inductive  kick  from  which  breaks  down  the  high  initial  resist- 
ance of  the  tube. 

Because  of  its  low  power  factor,  the  use  of  the  alternating-cur- 
rent mercury  vapor  lamp  is  objectionable. 

7.  The  quartz  lamp.  —  The  fundamental  principle  of  the  quartz 
lamp  is  the  same  as  that  of  the  mercury  vapor  lamp  in  that  its  light 
is  due  to  luminescent  mercury  vapor.     The  pressure  of  the  mercury 
vapor  in  the  quartz  lamp  is  materially  higher  than  in  the  mercury 
vapor  lamp,  the  temperature  and  the  luminous  intensity  are  cor- 
respondingly greater,  and  the  light  is  not  entirely  devoid  of  red  rays. 

The  quartz  lamp  emits  a  considerable  quantity  of  ultra-violet 
rays  which  are  decidedly  harmful  to  living  organisms,  and  should 
not  be  used  unless  it  is  enclosed  by  a  protecting  globe. 

8.  The  Moore  tube.  —  An  application  of  the  Geissler  discharge 
to  commercial  lighting  is  made  by  means  of  the  Moore  tube,  which 


ELECTRIC  LAMPS  243 

consists  of  a  highly  exhausted  glass  tube,  of  any  length  up  to  about 
200  feet,  the  luminous  properties  of  which  are  due  to  an  electric 
discharge  through  rarified  gases  introduced  into  the  tube.  The 
quality  of  the  light  produced  varies  with  the  medium  through  which 
the  electric  discharge  takes  place;  carbon  dioxide  gives  a  light 
approximating  daylight,  nitrogen  an  orange  tinted  light,  air  a 
pinkish  light. 

The  operation  of  the  tube  depends  essentially  on  the  maintenance 
of  the  proper  pressure  of  the  rarified  gas  in  the  tube,  the  electric 
discharge  forming  a  solid  precipitate  which  reduces  the  pressure  in 
the  tube.  As  the  pressure  decreases  the  conductivity  increases,  and 
the  increased  current  operates  a  valve  which  admits  gas  to  the  tube, 
and  restores  the  required  pressure. 


CHAPTER  XVI 
CIRCUIT-INTERRUPTING  APPARATUS 

THE  circuit-interrupting  apparatus  of  an  electric  plant  is  the 
means  by  which  the  generators  and  the  load  are  connected  and  dis- 
connected, and  includes  not  only  the  switches,  but  fuses  and  other 
circuit-breaking  devices,  and  the  auxiliary  apparatus  for  their  man- 
ipulation. The  satisfactory  operation  of  a  power  plant  depends, 
in  no  small  degree,  on  the  proper  selection,  installation,  and  opera- 
tion of  the  switching  gear  and  the  protective  apparatus. 

1.  Fuses.  —  A  fuse  is  a  short  piece  of  lead  and  tin  alloy,  of  such 
cross  section  and  so  connected  in  the  circuit  that  it  is  melted  and 
the  circuit  opened  when  an  excessive  current  flows.     Fuses  are  either 
"open''  or  " enclosed."     Enclosed  fuses  are  to  be  preferred  since 
the  arc  and  the  hot  metal  incident  to  the  opening  of  the  circuit  are 
confined,  thus  reducing  the  fire  hazard. 

The  current  at  which  a  fuse  will  open  a  circuit  can  be  determined 
only  approximately  because  of  external  conditions,  such  as  the 
temperature  of  the  air,  area  of  contact,  etc. 

2.  Switches.  —  Switches  are  for  the  specific  purpose  of  con- 
necting and  disconnecting  the  generator  and  the  load  apparatus, 
and  may  be  divided  into :  (a)  air-break  switches,  (b)  carbon-break 
switches,  (c)  oil-break  switches. 

(a)  Air-break  switches.  —An  air-break  switch  consists  of  one  or 
more  blades  of  copper  hinged  at  one  end  and  making  contact  at  the 
other  end  with  spring  clips  which  form  part  of  the  circuit.     As 
operating  switches,  i.e.,  for  opening  current-carrying  circuits,  air- 
break  switches  are  used  on  small  apparatus  only,  because  of  the 
burning  of  the  contacts  when  the  switch  is  opened;  for  completely 
isolating  other  apparatus  from  a  "live"  line,  they  are  universally 
used. 

(b)  Carbon-break  switches.  —  To  protect  the  copper  terminals  of 
an  air  break  switch,  a  circuit  having  carbon  terminals  is  connected 
in  parallel  with  the  one  having  copper  terminals.     In  operating  the 
switch,  the  copper  terminals  open  first,  without  an  arc,  thus  shunt- 

244 


CIRCUIT-INTERRUPTING  APPARATUS 


245 


ing  the  current  through  the  carbon  circuit  and  protecting  the  copper 
contacts  from  injury.  The  carbon  terminals,  which  may  be  re- 
newed, are  not  so  easily  burned  as  are  copper  terminals. 

(c)  Oil-break  switches.  —  For  the  manipulation  of  high  voltage 
circuits  or  those  carrying  large  currents,  switches  having  their  con- 
tacts immersed  in  oil  are  always  used. 
The  arc  formed  when  the  contacts  are 
separated  is  promptly  smothered  by 
the  oil  without  appreciable  damage  to 
the  contacts. 

Since  a  fuse  cannot  be  depended  on 
to  open  a  circuit  promptly  when  the 
current  exceeds  a  specified  value,  and 
must  be  renewed  each  time  it  operates, 
automatically  opening  switches  have 
been  devised  for  the  protection  of  elec- 
trical apparatus.  An  automatic  switch 
is  closed  against  the  pressure  of  a  spring, 
and  is  held  by  a  latch.  When  the  cur- 
rent exceeds  a  predetermined  value,  an 
electromagnet  trips  the  latch,  and  the 
spring  opens  the  switch. 

The  coils  of  the  tripping  magnet  are 
excited:  (i)  by  the  line  current,  the 
winding  of  the  magnet  being  connected  in  series  with  the  load, 
(2)  from  an  auxiliary  source,  the  circuit  through  the  coils  of  the 
magnet  being  closed  by  a  relay  connected  in  the  secondary  circuit 
of  a  series  transformer,  as  indicated  in  Fig.  205.  The  first  method 
is  applicable  to  either  continuous-  or  alternating-current  circuits; 
the  second,  to  alternating-current  circuits  only.  The  current  at 
which  an  automatic  switch  opens  is  adjusted  by  changing  the  length 
of  the  air  gap  in  the  magnetic  circuit  of  the  electromagnet,  or  the 
relay,  or  by  changing  the  tension  of  a  spring. 

In  many  cases,  particularly  in  motor  operation,  both  automatic 
switches  and  fuses  are  placed  in  the  circuit.  The  fuses  are  rated 
slightly  higher  than  the  current  at  which  the  switch  is  set  to  operate, 
and  their  purpose  is  to  protect  the  motor  in  case  the  operating 
mechanism  of  the  switch  becomes  deranged. 

By  the  addition  of  an  air  dash-pot  or  other  mechanical  device, 


FIG.  202.  Triple-pole,  Single-throw 
Non-automatic  Oil  Switch.  Gen- 
eral Electric  Co. 


246 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


the  tripping  mechanism  of  an  automatic  switch  is  made  less  sensi- 
tive to  momentary  overloads  or  surges,  i.e.,  the  switch  opens  only 

after  the  overload  has  been  maintained 
for  a  definite  length  of  time,  which 
may  be  made  inversely  proportional 
to  the  overload.  Such  devices  are 
known  as  time  limit  relays. 

3 .  Switch  operation.  —  Switching 
operations,  in  small  plants  may  be  done 
manually,  but  in  larger  plants  hand 
operation  becomes  unsatisfactory  and 
often  impossible,  because  of  the  size 
of  the  moving  parts,  as  well  as  danger- 
Section  through  Typi-  ous,  because  of  high  voltages.  Elec- 
trical power  is,  naturally,  used  for 
the  operation  of  large  switches.  The 


FIG.  203. 

cal  Bus  Bar  and  Switch  Com- 
partments.    Westinghouse. 


two  methods  of  switch  operation  in  general  use  are:  (a)  solenoid, 
(b)  motor. 

(a)  Solenoid-operated  switches.  —  Solenoid-operated  switches  re- 
quire the  use  of  two  solenoids,  one  for  closing  the  switch,  the  other 


D.  C.  Buses 


Red  Lamp  lighted 
when  oil  srv.  is  close 


Closing  Co  itact 
<A  Operating 


Opening"  Green  lamp  lighted 
Con  tact   "nen  oil  sw.  is  ope/ 
Terminal  board  on  oil  sw. 

Closing  Coil 


<•  C/ot  ed  when  oil  sty. 
is  closed 
when  oil  sw. 


FIG.  204.  Diagram  of  Connections  for 
Non-automatic  Solenoid-operated  Oil 
Switch. 


5rffen  Lamp-light* 
ft  hen  Oil  SH.  is  open  \ 
^"Terminal  board  on  \ 
oil  Switch 

Oyer  load  ReTab 
Closed  when  OilSw.  is  c.  osed 


'Ground 


Closed  when  oil''Sm 

(o^/s  open^Coil 

f-^^v    \^'Closed 'except when     , 
,-,  ',      T~~*     ^  '     •       motor  is  running 

Closed  only  when  motor  is 


running 


FIG.  205.  Diagram  of  Connections  for 
Automatic  Motor  -  operated  Oil 
Switch. 


for  opening  it.  In  alternating-current  plants  the  solenoids  (Fig. 
204)  are  usually  supplied  with  continuous  current  from  the  exciter 
buses,  and  are  energized  only  during  the  time  the  switch  is  opening 
or  closing.  When  the  switch  is  to  open  automatically,  it  may  be 
closed  against  the  pressure  of  a  spring,  and  the  opening  coil  replaced 


CIRCUIT-INTERRUPTING  APPARATUS 


247 


I 

•s 
£ 


J 


248 


ESSENTIALS  OF   ELECTRICAL  ENGINEERING 


CIRCUIT-INTERRUPTING   APPARATUS 


249 


by  a  tripping  coil  which  releases  a  latch,  and  allows  the  spring  to 
open  the  switch. 

(b)  Motor-operated  switches.  —  In  motor-operated  switches,  the 
actual  closing  and  opening  of  the  switch  is  done  by  springs,  the 
function  of  the  motor  being  to  wind  up  the  springs  after  each  opera- 
tion. Closing  the  circuit  of  a  tripping  coil  or  control  magnet,  auto- 
matically or  by  hand,  operates  the  switch,  the  movement  of  the 
switch  parts  closing  the  circuit  of  the  motor  so  that  it  winds  up  the 
springs  ready  for  the  next  operation.  Fig.  205. 

Bull's  eye  lamps,  usually  red  and  green,  the  circuits  of  which  are 
opened  and  closed  by  the  movement  of  the  switch  parts,  indicate 
whether  the  switch  is  open  or  closed. 


FIG.  208.     Typical  Benchboard.     General  Electric  Co. 

4.  Switchboards.*  -  -  The  switching  apparatus  and  the  indicating 
instruments  of  a  plant  are  grouped  and  placed  on  slate  or  marble 
panels  which  are  combined  to  form  the  switchboard,  typical  ex- 
amples of  which  are  shown  in  Figs.  206  and  207.  The  arrange- 
ment of  the  apparatus  is  a  matter  of  convenience,  indicating  instru- 
ments being  placed  near  the  top  of  the  board  where  they  are  easily 
read;  switch  and  rheostat  handles  at  such  a  height  as  to  be  easily 

*  For  details  of  switchboard  construction  the  reader  is  referred  to  publications  of 
manufacturing  companies. 


2  5° 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


Outgoing  Lines 
L*£- 


'  Lightning  Arresters 
feeder  GK  Be. 

1 r-4 


Uv     v4#     «L     ^     4v 

' 


L.T.BUS. 


Dis.Sw. 
Dis.Sw. 


FIG.  209.     Skeleton  Wiring  Diagram  (Single  Line). 


To  Feeder 


-^-  Tronsforqrter  Ground 

Ground  Bus 

To  Feeder  Jo  Feeder 


KEY  TO  SYMBOLS 

A  =  Ammeter.  P.T.       =  Potential  Transformer. 

A.S.       =  Three-way  ammeter  switch.  P.R.W '.  —  Polyphase  watthour  meter. 

T.B. 


B,AS.  =  Bell  alarm  switch. 

C.T.  =  Current  transformer. 

F  =  Fuse. 

O.S.  =  Oil  switch. 

FIG.  210.    Typical  Wiring  Diagrams  for  Three-phase  Feeder  Panels.     General 

Electric  Co. 


Terminal  board  for  secondary 
leads  from  current  and  po- 
tential transformers. 
T.C.       =  Trip  coil  on  oil  switch. 


CIRCUIT-INTERRUPTING   APPARATUS  251 

operated;  recording  meters  and  other  apparatus  requiring  only 
periodic  or  infrequent  attention,  near  the  floor,  and  on  either  the 
front  or  the  back  of  the  panel. 

Electrical  operation  of  switches  permits  of  maximum  concen- 
tration of  control  apparatus.  Control  wiring  involves  only  low 
voltage  circuits  of  small  ampere  capacity.  The  wires  and  apparatus 
are,  therefore,  of  small  size  and  may  be  placed  close  together. 
Control  levers,  indicating  lamps,  instruments,  relays,  etc.,  may  be 
placed  on  a  " bench"  or  control  board  (Fig.  208)  the  location  of 
which  is  entirely  independent  of  the  location  of  the  switches. 


CHAPTER  XVII 
METERS  * 

i.  Ammeters  and  voltmeters.  —  Commercial  ammeters  and 
voltmeters  operate,  in  general,  on  the  same  essential  principles. 
The  current- carrying  coils  of  an  ammeter  consist  of  a  few  turns  of 
heavy  wire  connected  in  series  with  the  load  apparatus;  the  cur- 
rent-carrying coils  of  a  voltmeter  consist  of  a  large  number  of  turns 
of  fine  wire  connected  in  parallel  with  the  load  apparatus.  Volt- 
meter and  ammeter  connections  are  shown  in  Fig.  211. 

Ammeters  and  voltmeters  may  be  classified  as:  (a)  hot  wire 
instruments,  (b)  permanent  magnet  instruments,  (c)  electrodyna- 
mometers,  (d)  soft  iron  instruments,  (e)  induction  instruments. 

(a)  Hot  wire  instruments.  —  The  fact  that  an  electric  current 
heats  the  conductor  through  which  it  flows,  and  that  the  length  of 
the  conductor  varies  with  the  temperature,  are  made  use  of  in  the 
design  of  hot  wire  instruments.  The  operation  of  such  an  instru- 


6hunt 


FIG.  211.    Voltmeter  and  Ammeter        FIG.  212.     Diagram  of  Hot  Wire 
Connections.  Instrument. 

ment  will  become  clear  from  a  study  of  Fig.  212.  Since  the  cur- 
rent-carrying wire  is  always  under  tension,  it  tends  to  assume  a 
permanent  "set"  which  destroys  its  calibration. 

(b)  Permanent  magnet  instruments.  —  In  this  type  of  instrument 
a  current-carrying  coil  is  pivoted  between  the  poles  of  a  permanent 
magnet.  When  current  flows  in  the  coil,  the  reaction  between  the 
magnetic  flux  and  the  coil  produces  a  torque,f  and  the  coil  tends 

*  This  discussion  of  meters  is  necessarily  very  brief,  but  the  writer  believes  that 
even  this  short  study  will  be  found  both  interesting  and  profitable.  For  a  detailed 
discussion  of  the  design,  construction  and  operation  of  electrical  meters  and  their  aux- 
iliary apparatus,  the  reader  is  referred  to  "  Electrical  Meters,"  by  Cyril  M.  Jansky. 

t  See  Chapter  2,  Section  14. 

252 


METERS 


253 


to  rotate,  but  its  movement  is  opposed  by  springs  and  the  coil 
assumes  that  position  where  the  torque,  which  is  proportional  to 
the  current  flowing  in  the  coil,  is  balanced  by  the  tension  of 
the  springs.  Because  of  its  permanent  magnetic  field,  this  type 
of  instrument  is  applicable  to  continuous-current  circuits  only. 

The  movable  coil  is  often  wound  over  an  aluminum  frame 
(Fig.  213)  which  acts  as  a  supporting  structure,  and  also  tends  to 
make  the  instrument  "dead  beat,"  i.e.,  to  cause  the  coil  to  come  to 
rest  without  a  prolonged  series  of  oscillations,  the  movement  of 
the  aluminum  frame  in  the  magnetic  field  inducing  currents  in 
the  frame,  which  oppose  its  movement.* 

(c)  Electrodynamometer.  —  The    electrodynamometer    is    similar 
in  operation  to  permanent  magnet  instruments,  the  magnetic  field 
being  produced  by  a  stationary  coil  con- 
nected in  series  with  a  movable  coil.     Be- 
cause   of    the    varying    strength    of    the 

magnetic  field,  the  deflection  of  the  mov- 
able coil  is  proportional  to  the  square  of 
the  current,  and  the  divisions  of  the  scale 
are  not  uniform.  The  electrodynamometer 
is  applicable  to  either  continuous  or  alter- 
nating-current measurements. 

(d)  Soft    iron    instruments.  —  Soft    iron 

. "  i    j        /  \    ,1          i  FIG.  213.     Operating  Prin- 

instruments  include:  (i)  the  plunger  type,     ciple  of  permanent  Mag- 
in  which  a  fixed  current-carrying  coil  and  a 
movable  soft  iron  plunger  react  to  move  a 
pointer  over  a  scale  as  indicated  in  Fig.  214,  (2)  the  magnetic  vane 
type,  in  which  the  movable  coil  of  the  electrodynamometer  is 
replaced  by  a  vane  of  soft  iron  to  which  the 
pointer  is  attached.     Soft  iron  instruments  are 
cheap,    and    applicable    to    either    continuous^ 
or    alternating-current    circuits,    but    are    less 
FIG.  214.     Schematic  acc^rate  than  other  types. 
Diagram  of  Plunger       (e)  Induction  instruments.  —  When  a  disc  or 
Ammeter  (or  Volt-  drum  of  copper,  aluminum  or  other  conducting 
material  is  placed  between  the  poles  of  an  alter- 
nating-current  magnet,   currents   are   induced   in    the   metal   by 
the  changing  flux.f    By  the  addition  of   a  "shading"    coiljj   as 

*  See  Chapter  2,  Section  13.  f  See  Chapter  2,  Section  13.  J  See  Chapter  13,  Section  17. 


net  Ammeters  and  Volt- 
meters. 


Plunger 


•Pointer 


254 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


shown  in  Fig.  2i7a,  or  of  a  transformer  coil  and  an  auxiliary 
winding,  as  indicated  in  Fig.  2170,  a  shifting  flux  is  produced, 
and  the  reaction  between  this  shifting  flux  and  the  currents 


FIG.  215.  Edgewise  Permanent  Magnet 
Voltmeter.  Westinghouse  Electric  & 
Mfg.  Co. 


FIG.  216.  Soft  Iron  Core  Ammeter. 
(Without  Case.)  Westinghouse 
Electric  &  Mfg.  Co. 


A.C.    Line 


induced  in  the  disc  or  drum,  causes  the  deflection  of  the  latter 
against  the  tension  of  a  spring.*  Induction  instruments  are  ap- 
plicable to  alternating-current  circuits  only. 

2.  Ammeter  shunts.  —  If 
the  entire  current  output  of 
a  large  plant  were  to  flow 
in  the  coils  of  an  ammeter, 
the  meter  would  be  both 
expensive  and  cumbersome. 
If  the  meter  coil  is  connected 
in  parallel  with  a  resistance, 


CcO 


(b) 


FIG.  217.     Diagram  of  Connections  for  Induc- 
tion Ammeter. 


as  indicated  in  Fig.  2i8a,  the 
currents  in  the  two  branches 

are   inversely   proportional    to    the  resistances  of   the  branches, 

and  the  ammeter  indication  is  proportional  to  the  total  current  in 

the  circuit. 

Such   a  resistance,  known  as  an  ammeter  or  current  shunt,  is 

used  with  practically  all  commercial  ammeters  intended  for  use 

*  See  Chapter  13,  Section  17. 


METERS 


255 


A.C.   Line 


D.  C.    L  ine 


on  continuous-current  circuits.  On  instruments  indicating  twenty- 
five  amperes  or  less,  the  shunts  are  enclosed  in  the  cases  of  the 
instruments;  for  instruments  of  greater  capacity,  external  shunts 
are  usually  used.  Fig.  219. 

3.  Series  transformers.  — 
On  alternating-current  circuits, 
the  ammeter  shunt  is  replaced 
by  a  series  or  " current"  trans- 
former. If  the  primary  wind- 
ing of  a  transformer  is  con- 
nected in  series  with  the  load, 
as  indicated  in  Fig.  2i8b,  a 
certain  fall  of  potential  takes 
place  when  current  flows  in  the 
windings.  This  voltage  drop  is  due  to  the  impedance  of  the  wind- 
ings, and  is  directly  proportional  to  the  current,  the  impedance 
of  the  windings  being  constant.  Since  there  is  a  constant  ratio* 
between  the  primary  and  the  secondary  voltages  of  a  transformer, 
the  secondary  voltage  is  proportional  to  the  current  flowing  in  the 
primary  circuit.f  Therefore,  if  the  impedance  of  the  secondary 
circuit  is  constant,  the  secondary  current  is  proportional  to  the 
primary  current. 


Cb) 


FIG.  218.     Series  Transformer  and  Ammeter 
Shunt  Connections. 


6/-0S+/ 


FIG.  219.    Ammeter  (Current)  Shunt.     Wagner  Electric  Mfg.  Co. 

Series  transformers,  when  designed  for  use  with  ammeters  or 
wattmeters,  are  usually  so  proportioned  that  the  full-load  secondary 
current  is  five  amperes.  If  the  instrument  is  not  calibrated  to 
read  direct,  the  instrument  indication  must  be  multiplied  by  the 
ratio  of  current  transformation. 

*  See  Chapter  12,  Section  3. 

t  Warning.  —  If  the  primary  current  remains  constant,  the  secondary  voltage  of  a 
series  transformer  increases  with  the  impedance  of  the  circuit  and  this  type  of  trans- 
former should  never  be  operated  with  the  secondary  circuit  open.  Before  opening  any 
circuit  supplied  from  the  secondary  winding  of  a  series  transformer,  short-circuit  the  sec- 
ondary terminals  of  the  transformer. 


256 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


•  |  Multiplier 


(        J  Voltmeter 
(a)  <b> 

FIG.  220.     Voltmeter  Connections,    (a)  Multi- 
plier.    (6)  Transformer. 


4.  High  voltage  measurements.  —  The  above  types  of  instru- 
ments are  not  adapted  to  the  direct  measurement  of  voltages 
above  seven  hundred  volts.  For  measuring  voltages  higher  than 
this  there  must  be  used:  (a)  a  multiplier,  (b)  a  potential  trans- 
former, (c)  a  special  voltmeter. 

(a)  The  multiplier.  —  A  multiplier  is  a  resistance  connected  in 
series  with  the  current-carrying  coil  of  a  voltmeter,  or  the  volt- 
meter   element    of    a    watt- 
meter, and  serves  simply  to 
reduce  the  voltage  that  would 
otherwise  be  applied  to   the 
terminals  of   the  coil.      The 
resistance   of   the   multiplier 
should  bear  a  simple  ratio  to 
the    resistance    between    the 
terminals    of   the    voltmeter 
with  which  it  is  to  be  used, 
so    that   the   voltage   of  the 

circuit  is  some  multiple  of  the  meter  indication.  The  multiplier 
is  applicable  to  continuous-  or  to  alternating-current  circuits. 
Fig.  22oa. 

(b)  The  potential  transformer.  —  The  voltage  of  an  alternating- 
current    circuit    may    be    stepped  Po.nter 

down  by  means  of  a  small  potential 
transformer,  the  voltage  of  the 
primary  circuit  being  the  indication 
of  the  voltmeter  multiplied  by  the 
ratio  of  transformation.  Fig.  22ob. 

(c)  Special  voltmeters. — The  volt- 
age   of    a    high  potential    system 
is  determined  by  means  of:  (i)  the 
.electrostatic    voltmeter,     (2)     the 
spark  gap. 

(i)  The  electrostatic  voltmeter.  — 
The  principle  on  which  the  electro- 
static voltmeter  operates  is  the  attraction  between  two  oppositely 
charged  bodies.  *  The  essential  parts  of  the  Westinghouse 
electrostatic  voltmeter  are  shown  schematically  in  Fig.  221.  A  A 

*  See  Appendix  C,  Section  2. 


FIG.  221.  Schematic  Diagram  of  Con- 
nections for  Westinghouse  Static 
Voltmeter. 


METERS 


257 


are  curved  metallic  plates  connected,  through  condensers,  to 
the  conductors,  the  difference  of  potential  between  which  it  is 
desired  to  measure;  and  BB  are  hollow  cylinders  to  which  a 
pointer  is  attached.  The  position  of  the  hollow  cylinders  changes 
as  the  voltage  between  lines  changes,  and  the  scale  is  calibrated  to 
read  volts. 

2&  26  24  22  20  18  16  14 


270 


240 


210 


180 


120 


90 


30 


\ 


4Ss 


I 


10 


IZ 


14 


2466 
INCHES 
FIG.  222.    Sparking  Distances. 

Standard  Conditions:  oo  sewing  needles. 
25°  Centigrade. 
760  mm.  barometer. 
&o%  humidity. 
Sinusoidal  voltage  wave  form. 

(2)  The  ^>ar&  gap.  —  The  voltage  required  to  break  down  the 
resistance  of  the  air  between  two  needle  points  is  approximately 
constant.  If  the  distance  between  points  at  which  the  voltage  of 
a  given  line  breaks  down  the  intervening  air  is  determined,  the 
voltage  of  the  line  may  be  determined  by  reference  to  a  table  or  a 
curve.  Fig.  222. 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


5.  Wattmeters.  —  A  wattmeter  is  a  combination  of  a  voltmeter 
and  an  ammeter  operating  on  the  same  movable  element,  so  that 
the  deflection  of  a  pointer  is  proportional  to  the  product  of  the 
current  in  the  circuit  and  the  voltage  between  the  terminals  of 
the  voltage  coil.  Two  types  of  wattmeters  have  been  developed: 
(a)  electrodynamometer,  (b)  induction. 

(a)  Electrodynamometer  wattmeter.  —  The  electrodynamometer 
wattmeter  consists  of  a  fixed  coil  of  large  wire  (ammeter  element) 

connected  in  series  with  the  load,  and  a 
movable   coil    of   fine   wire    (voltmeter 
element)  connected  in  parallel  with  the 
load.     The  ammeter  element  sets  up  a 
flux  with  which  the  voltmeter  element 
reacts  to  cause  a  deflection  of  the  fine 
FIG.  223.     Schematic  Diagram  wjre   Coil,*   and   the    deflection   is  pro- 
of Dynamometer  Wattmeter.       portional  to  ^  product  of  ^  currents 

in  the  coils,  i.e.,  to  the  power  in  the  circuit.  This  type  of  watt- 
meter indicates  the  power  in  either  continuous-  or  alternating- 
current  circuits.  Fig.  223. 


Line 

Movable  (Shunt) 

<\  c°!l 

Fixed  (.Series) 

Coils 

FIG.  224.    Weston  Wattmeter  (Cover 
Removed). 


FIG.  225.     Movable  Element  of  Weston 
Wattmeter. 


(b)  Induction  wattmeter.  —  The  induction  wattmeter  is  identical 
in  principle  with  the  induction  ammeter  and  voltmeter  described 
above,  i.e.j  deflection  of  a  pointer  is  caused  by  the  reaction  between 
a  shifting  flux  and  the  currents  induced  in  a  disc  or  drum.  The 
ammeter  element,  which  is  wound  with  a  few  turns  of  heavy  wire, 
has  negligible  inductance;  the  voltmeter  element,  which  is  wound 
with  many  turns  of  fine  wire,  has  large  inductance.  The  relations 

*  See  Chapter  2,  Section  14. 


METERS  259 

of  the  fluxes  are  as  represented  in  Fig.  226a,  i.e.,  the  flux  due  to 
the  parallel  winding  lags  behind  that  due  to  the  series  winding 
by  the  angle  0. 

The  angle  0  in  Fig.  226a  is  increased  to  90  degrees  by  means  of 
an  auxiliary  or  compensating  winding  which 
is    a    short-circuited    winding    similar    to    a 
"shading  coil,"  *  except  that  it  surrounds  the 
entire  pole.     The  compensating  coil  acts  as 
the  short-circuited  secondary  of  a  transformer, 
and  sets  up  a  flux  the  magnitude  and  phase     i 
relations  of  which  are  represented  by  OC  in      ||%^^5H^ 
Fig.  226b.     The  flux  in  the  magnetic  circuit     ^| 
of  the  parallel  winding  is,  then,  the  geometric      If  * .;--     ^uxdueto  shunt 

5         /AS         :       B     Winding 

difference  of  that  due  to  the  shunt  coil  and      \$^£3v 

,-,  r          ,1  ,•  •      j.  C*-"""" ''Resultant  Flux. 

that   set   up  by  the  compensating  winding.  ^ 

By  properly  proportioning  the  compensating  FIG.  226.  Vector  Dia- 
winding,  a  flux  in  exact  quadrature  with  that  grams  of  Fluxes  in  the 
set  up  by  the  series  winding  is  produced.  Induction  Wattmeter. 
The  resultant  of  these  quadrature  fluxes  is  a  shifting  or  rotating 
flux,  and  a  torque  which  is  proportional  to  the  product  of  the 
voltage  of  the  circuit  and  the  load  amperes.  Like  other  induction 
apparatus,  the  induction  wattmeter  is  applicable  to  alternating- 
current  circuits  only. 

That  a  properly  designed  wattmeter  indicates  the  power  in  an 
alternating-current  circuit  when  the  load  is  either  non-inductive 
or  inductive  is  evident  when  the  torque  relations  during  a  cycle 
aje  considered.  In  a  non-inductive  single-phase  circuit,  the  torque 
is  constant  in  direction  but  varies  in  value  from  zero  to  maximum, 
and  the  wattmeter  indication  is  proportional  to  the  average  torque. 
In  an  inductive  circuit,  the  direction  of  the  torque  is  not  constant, 
and  the  wattmeter  indication  is  proportional  to  the  algebraic  sum  of 
the  average  positive  and  negative  torques.  The  current  and  volt- 
age remaining  constant,  it  may  be  proved  both  mathematically  and 
experimentally,  that  the  average  net  torque  is  proportional  to  the 
cosine  of  the  phase  angle.f  The  deflection  of  the  movable  system 
of  a  wattmeter  is,  therefore,  proportional  to  the  product  of  the 
current,  the  electromotive  force,  and  the  cosine  of  the  angle  by 

*  See  Chapter  13,  Section  17. 
t  See  Chapter  i,  Section  28. 


260  ESSENTIALS   OF   ELECTRICAL   ENGINEERING 

which  the  current  leads  or  lags  behind  the  electromotive  force,  i.e., 
to  the  power  in  the  circuit. 

6.  Polyphase  wattmeters.  —  For  the  measurement  of  power  in 
polyphase  alternating-current  circuits,  polyphase  wattmeters  have 
been  designed.     The  polyphase  wattmeter  is  a   combination  of 
two  or  more  single  wattmeter  elements  acting  on  the  same  movable 
part,  and  each  pair  of  coils  is  connected  as  if  it  formed  an  inde- 
pendent instrument. 

7.  Watt-hour  meters.  —  A  watt-hour  meter  is  a  small  motor, 
the  speed  of  the  rotating  parts  of  which  is  proportional  to  the  power 
in  the  circuit  to  which  it  is  connected.     The  movable  parts  of  such 
motors  are  made  very  light,  and  friction  is  reduced  to  a  minimum 
by  the  use  of  jewel  bearings. 

Watt-hour  meters  are  divided  into  two  classes:  (a)  commutator 
meters,  (b)  induction  meters. 

(a)  Commutator  meters.  —  Commutator  watt-hour  meters  are 
similar,  in  principle  and  in  action,  to  the  continuous-current  shunt 
motor,  and  are  applicable  to  either  direct-  or  alternating-current 
circuits.  The  armature  is  wound  of  many  turns  of  fine  wire  and 
is  connected  in  series  with  a  resistance,  between  the  supply  lines. 
The  field,  which  is  produced  by  the  series  windings,  is  propor- 
tional to  the  load  current,  the  magnetic  circuit  being  in  air. 

With  constant  electromotive  force  between  the  terminals  of 
the  armature  winding,  the  torque  of  the  motor  is  proportional  to 
the  current  flowing  in  the  field  windings.  Therefore,  to  make  the 
speed  proportional  to  the  power  in  the  circuit,  the  load  on  the  motor 
(counter  torque)  must  be  proportional  to  the  speed.  This  is  ac- 
complished by  means  of  an  eddy-current  brake,  consisting  of  a  disc 
of  copper  or  aluminum  attached  to  the  shaft  of  the  motor,  and 
rotated  between  the  poles  of  a  permanent  magnet.  When  current 
flows  in  the  coils  of  the  meter,  a  torque  is  produced,  and  the  movable 
parts  of  the  meter  increase  in  speed  until  the  driving  torque  is 
balanced  by  the  retarding  torque  of  the  permanent  magnet  and  the 
eddy  currents  induced  in  the  rotating  disc.  Since  the  torque 
producing  motion  is  proportional  to  the  power  in  the  circuit  to 
which  the  meter  is  connected,  and  the  retarding  torque  is  propor- 
tional to  the  speed  of  the  rotating  parts,  the  speed  of  the  rotating 
parts  is  proportional  to  the  power  in  the  circuit.  By  means  of  a 
train  of  gears,  the  total  number  of  revolutions  made  by  the  arma- 


METERS 


261 


ture  is  recorded,  the  calibration  being  such  that  the  meter  reads  in 
either  watt-hours  or  kilowatt-hours.  For  a  given  load,  the  speed  of 
the  rotating  parts  of  a  meter  is  changed  by  changing  the  position 
of  the  permanent  magnet,  which  is  made  adjustable. 

Friction  between  the  moving  parts  of  the  meter  and  the  support- 
ing bearing,  which  would  disturb  the  torque-speed  relations  at 
light  loads  to  a  very  considerable  extent,  is  compensated  for  by  the 
addition  of  an  auxiliary  field 
winding  connected  in  series  with 
the  armature  winding,  and  so 
adjusted  that  when  zero  cur- 
rent flows  in  the  series  windings 
a  slight  jar  causes  the  armature 


Series  Field 


'Series  Field 


Schematic  Diagram  of  Con- 
nections for  Commutating  Watt-hour 
Meter. 


to  rotate.     If  the  field  set  up  by 

this    auxiliary    winding    is    too  FIG.  227. 

strong,  the  meter  " creeps,"  i.e., 

the  armature  rotates  when  the 

load  circuit  is  disconnected;  if  the  field  set  up  by  the  auxiliary 

winding  is  too  weak,  the  meter  does  not  register  on  light  loads. 
The  electrical  circuits  of  a  commutator  watt-hour  meter  are 

shown  in  Fig.  227. 

(ft)  Induction  meters.  —  The  induction  watt-hour  meter  is,  in 

principle,  a  two-phase  squirrel-cage 
induction  motor,  the  quadrature 
fluxes  being  produced  in  the  same 
manner  as  described  for  the  induc- 
tion watt-meter.  The  squirrel-cage 
element  of  the  induction  meter  is 
an  aluminum  disc  or  cup  which  also 
serves  as  the  movable  part  of  the 
eddy  current  brake,  thus  making 
the  movable  system  of  an  induction 

FIG.  228.    Schematic  Diagram  of  Con-    meter  lighter  than  that  of  the  COm- 
nections   for   Induction   Watt- meter  rr\  •  . 

and  Watt-hour  Meter.  mUtat°r  ^    The  tOr(!Ue  PCr  Umt 

weight  is  also  greater.  The  neces- 
sity for  friction  compensation  is,  therefore,  reduced,  but  is  provided 
by  means  of  a  " shading  coil"*  on  the  shunt  winding. 

An  induction  watt-hour  meter  is  shown  schematically  in  Fig.  228. 

*  See  Chapter  13,  Section  17. 


262 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


Because  of  the  principle  on  which  it  operates,  the  induction  watt- 
hour  meter  is  not  applicable  to  continuous-current  circuits. 

8.  Recording  or  graphic  meters.  —  It  is  often  desirable  to  have 
a  continuous  record  of  the  momentary  fluctuations  in  the  values  of 
such  electrical  quantities  as  voltage,  current,  power,  etc.  Such 
records  are  obtained  by  means  of  a  pen  or  pencil,  the  position  of 


SHUNT  COIL 


L*6HT  LCA&-* 
ADJUSTMENT 


MAGNET 


BALANCE 

LOOP 

DISK 


POWER  FACTOR 
/I  ADJUSTMENT 


FIG.  229.     Elements  of  Polyphase  Induction  Watt-hour  Meter. 
Westinghouse  Elec.  &  Mfg.  Co. 

which  is  controlled  by  the  magnitude  of  the  quantity  to  be  re- 
corded, and  a  uniformly  moving  paper  on  which  the  pen  or  pencil 
traces  a  record.  A  detailed  description  of  the  mechanism  of  such 
instruments  is  beyond  the  scope  of  this  work. 

9.  The  synchroscope.  —  The  synchroscope  or  synchronism  indi- 
cator is  a  device  by  means  of  which  the  phase  relations  of  two 
electromotive  forces  and  their  relative  frequencies  are  indicated. 

The  synchronism  indicator  used  by  the  General  Electric  Co. 
is,  structurally,  a  small  synchronous  motor  the  field  winding  of 
which  is  connected  to  the  bus  bars.  Fig.  230.  The  movable 
coils  are  connected  to  the  incoming  machine  through  a  phase- 


METERS 


263 


Movable 
Coils 


/ 


Resistance 


splitting  device.  There  is,  then,  superimposed  on  the  field  set 
up  by  the  movable  coils,  an  alternating  flux  due  to  the  station- 
ary winding,  and  the  movable  parts  assume  a  fixed  position  which 
is  dependent  on  the  phase  relation 
between  the  electromotive  force 
of  the  incoming  machine  and  that 
of  the  bus  bars,  the  frequencies 
of  the  two  electromotive  forces  be- 
ing the  same.  Correct  phase  rela- 
tions for  parallel  connection  of  al- 
ternators (phase  opposition)  exist 
when  the  stationary  ("  shadow  ") 
and  movable  pointers  coincide. 

If  the  frequencies  of  the  electro- 
motive  forces  are  not   the   same,    FIG.  230.    Schematic  Diagram  of  Con- 
a  torque  which  causes  the  arma-      nections  .for  General  Electric  synchr°- 

.  .  nism  Indicator. 

ture  to  rotate  is  produced.     The 

direction  of  rotation  is  clockwise  or  counter  clockwise  according  as 
the  speed  of  the  incoming  machine  is  too  fast  or  too  slow,  the  rate 
at  which  the  pointer  rotates  indicating  the  difference  in  the  fre- 
quencies of  the  electromotive  forces. 

The  Weston  synchro- 
scope operates  on  the  elec- 
trodynamometer  principle, 
the  movable  element  vibrat- 
ing instead  of  rotating.  By 
reference  to  Fig.  231,  it  will 
be  seen  that  the  fixed  coils 
are  connected  to  the  bus 
bars  in  series  with  a  slightly 
inductive  resistance,  and  the  movable  coil  to  the  incoming  machine 
in  series  with  a  condenser.  The  relative  values  of  the  capacitance, 
the  resistance  and  the  inductance  are  such  that  the  currents  are  in 
exact  quadrature  when  the  voltage  of  the  incoming  machine  is  in 
phase,  or  in  phase  opposition,  with  that  of  the  bus  bars.  Under 
this  condition  no  torque  is  exerted  between  the  coils,  and  the  pointer 
is  at  rest  in  the  middle  of  the  scale. 

When  the  electromotive  forces  are  not  in  phase  or  in  phase  oppo- 
sition, a  torque,  proportional  to  the  phase  displacement,  causes  the 


FIG.  231. 


Schematic  Diagram  for  Weston 
Synchroscope. 


264  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

movable  coil  to  be  deflected,  and  the  pointer  is  moved  to  the  right 
or  to  the  left  as  the  electromotive  force  of  the  incoming  machine 
leads  or  lags  behind  that  of  the  bus  bars. 

If  the  frequencies  are  not  the  same,  the  phase  displacement,  and 
consequently  the  position  of  equilibrium,  changes  momentarily, 
and  the  pointer  swings  back  and  forth  across  the  scale.  The 
lamp,  which  illuminates  the  dial  of  the  instrument,  is  so  connected 
that  it  is  dark  when  the  voltages  are  in  phase,  and  light  when  they 
are  in  phase  opposition.  Consequently,  the  pointer  seems  to  rotate 
either  clockwise  or  counter  clockwise,  the  direction  of  apparent 
rotation  indicating  whether  the  incoming  machine  is  too  fast  or  too 
slow,  and  the  rate  of  apparent  rotation,  the  amount  by  which  the 
frequencies  differ. 

10.  Power-factor  meters.  —  If  the  stationary  coils  of  a  synchro- 
scope are  connected  in  series  with  the  load  apparatus  and  the  mov- 
able coils  between  the  supply  mains,  the  position  of  the  movable 
element  depends  on  the  phase  relations  of  the  currents  in  the  two 
coils  as  explained  in  Section  9.  Since  the  phase  relations  of  the 
currents  in  the  coils  are  dependent  on  the  power  factor  of  the  load 
circuit,  the  scale  may  be  calibrated  to  read  power  factors. 

The  Weston  power-factor  meter,  —  The  movable  element  of  the 
Weston  power- factor  meter  differs  from  that  in  the  synchroscope 

in   that  it  has   two   coils,  the  planes 
of   which    are    at   right   angles.     The 
stationary  coils  are  connected  in  series 
with    the   load,   or   to   the   secondary 
terminals  of  a  series  transformer,  and 
the     movable    coils    across    different 
phases  of  a  polyphase  system  or  are 
FIG.  232.    Schematic  Diagram  of  provided  with  a  phase-splitting  device. 
Connections  for  Weston  Power-  Fig.  232.     If  the  current  in  the  sta- 
factor  Meter.  ^       tionary  coils    is    in    phase  with   that 

in  either  of  the  movable  coils,  the  plane  of  one  movable  coil 
coincides  with  the  axis  of  the  fixed  coils;  if  the  current  in 
the  fixed  coil  is  not  in  phase  with  that  in  either  of  the  mov- 
able coils,  the  movable  system  is  deflected,  and  the  angle  of 
deflection  is  equal  to  the  angle  between  the  current  in  the  sta- 
tionary coil  and  that  in  one  of  the  movable  coils,  i.e.,  to  the 
power-factor  angle. 


METERS' 


265 


Iron  Vane 


Westinghouse  power-factor  meter.  —  The  essential  parts  of  a 
Westinghouse  power-factor  meter  are:  (a)  a  soft  iron  vane  or 
armature  of  the  shape  indicated  in  Fig.  233,  (b)  two  or  more  angu- 
larly displaced  stator  coils  which  are  connected,  through  series 
transformers,  with  a  polyphase  sys- 
tem, (c)  a  stationary  potential  coil, 
the  axis  of  which  coincides  with  that 
of  the  iron  vane. 

The  stator  coils  set  up  a  rotating 
flux  which,  when  unaffected  by  other 
forces,  causes  the  armature  to  rotate.  FlG-  233-  Schematic  Diagram  of 

„«  ,.    ,        .,        ,      .         ,,  Westinghouse  Power- factor  Meter. 

The  potential  coil  polarizes  the  arma- 
ture and,  at  unity  power  factor,  it  assumes  a  position  at  right 
angles  to  the  stator  coil,  the  current  in  which  is  in  phase  with 
the  current  in  the  potential  coil.  When  the  power  factor  of  the 
system  is  not  unity,  the  current  in  the  potential  coil  is  not  in 
phase  with  that  in  the  stator  coil,  and  the  armature  is  deflected 
through  an  angle  equal  to  the  angle  of  phase  displacement. 

Polyphase  power-factor  meter.  —  The  indications  of  a  split-phase 
meter  are,  obviously,  affected  by  the  frequency  of  the  circuit  to 
which  it  is  connected.  This  difficulty  is  overcome  by  connect- 
ing the  movable  coils  to  a  polyphase  system,  thus  utilizing  the 
inherent  voltage  displacement  of  such  a  system.  The  operation  of 
such  an  instrument  is  in  no  other  way  different  from  that  of  the 

split-phase  meter. 

ii.   Frequency     meters.  — 
Frequency  meters  indicate  the 
frequency  at  which   a   system 
is  operating,  and  are  of   two 
types:     (a)    vibrating-reed 
meters,  (b)  split-phase  meters. 
FIG.  234.    Principle  of  Operation  of  Vibrat-      (a)   Vibrating-reed  meters.  — 
ing  Reed  Frequency  Meter.  Jf  a  ^^   of  ^  ^^  djf_ 

fering  in  length  and  in  inertia,  are  arranged  in  front  of  an 
alternating-current  magnet,  as  indicated  in  Fig.  234,  those 
strips  whose  natural  period  of  vibration  corresponds  to  the 
frequency  of  the  alternating-current  system  are  thrown  into 
violent  vibration,  while  the  others  are  affected  only  slightly  or 
not  at  all. 


Steel  Reeds 


.    \ 


*Solder 
Weight 


Electro-ma'qnet 


AC.  Supply 


266 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


(b)  Split- phase  meters.  —  Split-phase  frequency  meters  are 
essentially  differential  induction  voltmeters.  One  coil  is  con- 
nected to  the  mains  in  series  with  a  non-inductive  resistance  and 
its  current  is,  therefore,  practically  independent  of  frequency;  the 


A.     C. 

Line 

r 

Inductance 

Resistance  > 

I'l'l'l'l1 

Zl      22     23     «      25      26      27      26 

FIG.  235.  Section  of  Vibra ting-Reed  Fre- 
quency Meter  Scale  Showing  Method 
of  Indication. 


1 


Aluminum 
Disc  \ 


n 


FIG.  236.  Schematic  Diagram  of  Con- 
nections for  Induction  (Split-phase) 
Frequency  Meter. 


other  coil  is  connected  to  the  same  mains  through  an  inductive 
resistance  which  causes  the  current  to  vary  as  the  frequency 
changes.  At  normal  frequency,  the  torque  of  one  coil  is  equal  and 
opposite  to  that  of  the  other;  as  the  frequency  of  the  system  in- 
creases or  decreases  the  torques  are 
no  longer  equal,  and  equilibrium 
is  restored  by  a  deflection  of  the 
movable  element.  Fig.  236. 

In  the  Weston  frequency  meter 
two  stationary  coils  are  inter- 
connected with  resistances  and 
reactances  so  as  to  form  a  Wheat- 
stone  bridge  which,  at  normal 
frequency,  is  balanced.  As  the 
frequency  increases  or  decreases, 
the  bridge  is  unbalanced,  and  a 
movable  element  of  soft  iron 
caused  to  deflect.  The  electrical 


FIG.  237.     Westinghouse  (Split-phase) 
Frequency  Meter. 


connections  and  the  movable  element  of  this  meter  are  shown  in 
Fig.  238. 

12.  Ground  detectors.  —  A  "  ground  "  is  a  connection  between 
a  current-carrying  conductor  and  the  earth,  which  materially 
reduces  the  normal  insulation  resistance  of  the  line.  A  ground 


METERS 


267 


cannot  be  stated  to  exist  when  the  resistance  between  the  con- 
ductor and  the  earth  falls  below  a  certain  fixed  minimum,  because 
of  the  different  insulation  requirements  of  different  systems  — 
what  is  good  insulation  under  one  condition  may  be  a  very  serious 
ground  under  other  circumstances. 

A  ground  on  one  line  of  a  system  does  not,  in  itself,  cause  any 
trouble,  but  if  two  lines  become  grounded,  a  serious  leakage  occurs, 


Pointter 


Line 


FIG.  238.   Schematic  Diagram  for  Western 
Frequency  Meter. 


FIG.  239.   Schematic  Diagram  of  Static 
Ground  Detector. 


or  a  short-circuit  is  established.  Since  a  line  may  accidentally 
become  grounded  at  any  time,  it  is  good  practice  to  have  on 
the  switchboard,  instruments  which  indicate  the  fact  whenever  the 
resistance  between  any  line  and  the  earth  becomes  seriously  re- 
duced. Such  instruments  are  known  as  "ground  detectors." 

The  static  ground  detector  operates  on  the  same  principle  a.« 
the  electrostatic  voltmeter,  i.e.,  the  attraction  between  two  oppo- 
sitely charged  bodies.  Let  A  and  C  in  Fig.  239  be  curved  plates 
connected  to  current-carrying  conductors,  as  indicated;  B  a  plate 
which  is  free  to  move  about  the  pivot  and  which  is  connected, 
through  a  negligible  resistance,  to  the  earth.  If  the  lines  are 
equally  insulated,  equal  and  opposite  forces  tend  to  deflect  plate 
B  *  from  its  position  midway  between  plates  A  and  C.  If  the  lines 
are  not  equally  insulated,  the  equilibrium  of  forces  acting  on  B  is 
destroyed,  and  B  moves  toward  the  plate  (A  or  C)  which  is  con- 
nected to  the  line  having  the  higher  (better)  insulation,  and  the 
deflection  is  proportional  to  the  relative  resistances  between  the 
lines  and  the  earth. 

*  See  Appendix  C,  Section  2. 


CHAPTER  XVIII 
POWER  TRANSMISSION   AND   DISTRIBUTION* 

THE  advantages  of  power  transmission  by  means  of  the  electric 
current  are  such  that  this  method  has  practically  superseded  all 
other  methods  except  for  very  limited  distances.  Within  the  last 
few  years  a  number  of  large  hydroelectric  plants  have  been  com- 
pleted, the  power  from  which 
is  transmitted  one  hundred 
miles  or  more  at  voltages  up 
to  140,000.  The  transmission 
line  is  the  connecting  link  be- 
tween the  generators  and  the 
distributing  network,  and  is 
usually  run  over  a  private 
right  of  way,  the  current- 
carrying  conductors  being 
supported  by  wooden  poles 
or  steel  towers.  Fig.  240. 
Conductors  of  distributing 
systems  are  carried  on  poles 
or  are  placed  in  conduits 
underground. 

i.  Conductors.  -  -  Copper 
wires  or  cables  are  almost 
universally  used  for  the  trans- 
mission and  the  distribution 

of   electric    power,    although 
FIG.  240. 

aluminum   is   used   to   some 

extent  because  its  light  weight  makes  the  number  of  supporting 
structures  required  a  minimum.  The  sizes  of  commercial  conduc- 
tors are  expressed  by  their  areas  in  circular  mils,  or  by  gauge 

*  For  an  extended  discussion  of  transmission  problems,  the  details  of  pole  line  design, 
etc.,  the  reader  is  referred  to  "Overhead  Electric  Power  Transmission "  by  Alfred  Still, 
and  to  "Elements  of  Electrical  Transmission"  by  O.  ].  Ferguson. 

268 


POWER   TRANSMISSION  AND   DISTRIBUTION 


269 


numbers.  The  American  Wire  Gauge  (A.  W.  G.),  often  termed 
Brown  &  Sharpe  (B.  &  S.),  is  universally  used  in  the  United  States. 
A  characteristic  of  the  A.  W.  G.  which  makes  it  easy  to  ap- 
proximate the  area  corresponding  to  any  given  gauge  number 
without  the  use  of  a  wire  table,  is  the  fact  that  No.  10  is  approx- 
imately o.i  inch  in  diameter  and  has  an  area  of  approximately 
10,000  circular  mils,  and  that  the  area  halves  or  doubles,  approx- 
imately, for  each  three  gauge  numbers,  i.e.,  No.  7  has  an  approx- 
imate area  of  20,000  circular  mils  while  No.  13  has  an  approximate 
area  of  5000  circular  mils.  Gauge  numbers  and  areas  of  wires,  with 
their  weights  and  resistances,  are  given  in  Table  VI. 

2.  Insulation.  —  For  many  purposes  it  is  required  that  current- 
carrying  wires  be  covered  with  some  insulating  material.  The 
amount  and  the  quality  of  this  insulation  depends  on  the  use  for 
which  the  conductor  is  intended,  and  may  consist  of:  (a)  a  cotton 
braid,  (b)  a  cotton  braid  impregnated  with  a  liquid  compound 
which  dries  and  becomes  hard,  (c)  a  coating  of  vulcanized  rubber 
over  which,  as  a  mechanical  protection  for  the  rubber,  is  wound  a 
braid  having  a  polished  surface.  Rubber  insulation  is  required  for 
all  interior  wiring  by  a  rule  of  the  National  Electric  Code  (N.  E.  C.), 
which  has  been  adopted  by  the  national  engineering  societies  as 

well  as  by  the  National  Board  of 
Fire  Underwriters. 

Wires  and  cables  intended  for 
underground  use  are  thoroughly 
insulated,  then  covered  with  a 


(a)  Suspension  Insulator.  (b)  Pin  Insulator. 

FIG.  241.     Line  Insulators.     The  Ohio  Brass  Co. 

continuous  moisture-proof  lead  sheath  over  which  may  be  wound 
a  braid,  or  other  mechanical  protection. 

For  insulating  aerial  conductors  from  their  supporting  structures 
glass  or  porcelain  insulators  are  used.  Two  types  of  line  insulators 
are  in  general  use,  illustrations  of  which  are  shown  in  Fig.  241. 


270  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

Pin- type  insulators  are  generally  used  for  voltages  up  to  66,000; 
suspension  type  for  voltages  greater  than  this. 

3.  Carrying  capacity  of  conductors.  —  The   current  which  an 
insulated  conductor  may  safely  carry  is  limited  by  the  temperature 
at  which  the  insulation  softens,  or  by  reason  of  which  its  insulating 
properties   deteriorate.     The   maximum   current   allowed   by   the 
N.  E.  C.  is  given  in  Table  VII  for  rubber  and  for  other  insulations. 

4.  Inductance.  —  A  magnetic  field  which  opposes*  any  change  in 
the  value  of  the  current  is  set  up  around  any  current-carrying  con- 
ductor.f     Alternating-current  circuits  are,  therefore,  subject  to  an 
inductance  which  varies  with  the  distance  between  the  conductors 
and  with  their  size.     Inductances  for  different  sizes  of  wires  and  for 
different  spacings  are  given  in  Table  VIII,  the  values  being  those 
obtained  by  the  following  formula:  J 

2  D 
L  =  0.03028  +  0.282  logio  — —  >  (i) 

when  L  =  the  inductance  in  millihenries  per   1000  feet  of  two- 
wire  circuit  (2000  feet  of  conductor), 
D  =  the  distance  between  wires  in  inches, 
d  =  the  diameter  of  the  wires  in  inches. 

The  inductance  of  each  wire  in  a  three-phase  system  is  one-half 
that  of  the  loop  formed  by  any  two  of  the  three  conductors. 

5.  Capacitance.  —  Two  or  more  metallic  conductors  §  separated 
by  air  or  other  insulating  material,  form  a  condenser,  the  effect  of 
which  is   to   cause   a  leading  current  to  flow  in  an  alternating- 
current  circuit.     The  value  of  this  leading  current,  termed  the 
charging  current,  is  proportional  to   the  applied  voltage,  to  the 
capacitance  of  the  circuit,  and  to  the  frequency. 

Ie  (single  phase)  =  2  w/EC  io~6,  (2) 

/i                     \        4  irfEC  IO"6  f  N 

Ie  (three  phase)  =  *    J >  (3) 

V3 
when  E  =  the  line-to-line  voltage, 

C  =  the  capacitance  in  microfarads, 
/  =  the  frequency  of  the  supply  circuit, 
Ic  =  the  charging  current. 

*  Because  of  the  electromotive  force  induced  in  the  conductor  when  the  magnetic 
field  changes. 

t  See  Chapter  2,' Section  3.  J  See  Appendix  B,  Section  i.  §  See  Appendix  C, 
Section  9. 


POWER  TRANSMISSION   AND   DISTRIBUTION  271 

The  capacitance  of  a  transmission  line  varies  with  the  distance 
between  the  wires,  with  their  diameter,  and  with  the  nature  of  the 
insulating  material  between  the  conductors.  The  capacitances 
given  in  Table  IX  were  calculated  by  means  of  the  following  for- 
mula: * 

C-VS&1,  (4) 

loglo  — 

when  C  =  the  capacitance  in  microfarads  per  1000  feet  of  two-wire 

circuit  (2000  feet  of  conductor), 
D  =  distance  between  wires  in  inches, 
d  =  diameter  of  conductors  in  inches. 

6.  Skin  effect.  —  When  an  alternating  current  flows  in  a  con- 
ductor, the  current  density  over  the  cross-sectional  area  is  not  uni- 
form, but  increases  as  the  distance  from  the  center  of  the  conductor 
increases.     This  is  the  "skin"  effect,  and  increases  the  resistance 
losses  in  the  conductor.     The  increase  in  the  resistance  of  a  con- 
ductor to  alternating  currents  as  compared  with  its  resistance  to 
continuous  currents  is  a  function  of  the  product  of  the  area  of  the 
conductor   and   the   frequency   of   the   alternating   current.     The 
skin  effect  is  negligible  when  the  product  of  the  area  of  the  con- 
ductor in  circular  mils  and  the  frequency  does  not  exceed  30,000,000, 
or,  for  commercial  frequencies,  when  the  diameter  of  a  conductor 
is  not  greater  than  three-quarters  of  an  inch. 

7.  Line  calculations.  —  The  inductance  and  the  capacitance  of 
a  transmission  line  are  made  up  of  an  infinite  number  of  induc- 
tances and  capacitances  uniformly  distributed  over  the  system. 
An  exact  solution  of  such  a  system  involves  equations  too  compli- 
cated for  practical  use,  but  an  approximate  solution  giving  fairly 
accurate  results  is  easily  made  when  the  capacitance  is  assumed  to 
be  concentrated,  e.g.,  one-half  at  each  end  of  the  line.     Let 

EI  =  the  voltage  at  the  terminals  of  the  load, 
Ed  =  the  voltage  drop  in  the  line, 
E  =  the  voltage  at  the  terminals  of  the  generator, 
II  =  the  load  current, 

Ic  =  the  charging  current  due  to  the  capacitance  at  the  load 
end  of  the  line, 

*  See  Appendix  C,  Section  9. 


272  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

I  =  the  line  current, 
L  =  the  inductance  of  the  line, 
R  =  the  resistance  of  the  line, 
cos  <£  =  the  power  factor  of  the  load  circuit, 
<f>'  =  the  angle  between  EI  and  7, 


The  relations  between  Eh  It  and  7C  are  known,  and  their  vectors 
may  be  plotted  as  shown  in  Fig.  242,  and  the  value  of  7  and  that 
of  the  angle  <j>'  determined. 

^r   I  =  V7i2cos24>+(7,sin0-7c)2,   (5) 

(6) 


1 1  COS  0 

also, 

FIG.  242.  Vector  Diagram  for  Trans-     ^          T     /7^    T 
^  T  ,-r,  £Ld  =  1  V  K    -\- 


mission  Line.  * 

Adding,  vectorially,  the  line  drop  and  the  load  voltage,  the  volt- 
age at  the  generator  terminals  is  obtained. 

E  =  V[Et  +  Ed  cos  (e  -  0')]2  +  E2  sin2  (0  -  0')-  (8) 

Problem  —  Single  phase.  —  500  kw.  are  to  be  transmitted  a  dis- 
tance of  20  miles,  the  voltage  at  the  load  end  (primary  terminals 
of  step-down  transformers)  is  25,000,  the  frequency  is  25,  and  the 
power  factor  of  the  load  circuit  is  86.6  per  cent.  Find:  (a)  the 
drop  in  the  line,  (b)  the  voltage  at  the  generator  terminals. 

C  =  0.00155  X  5.28  X  20  =  0.1637  microfarad. 

Ic  =  2  TT  X  25,000  X  0.1637  X  25  X  10  =  0.64  ampere. 

L  =  0.694  X  5.28  X  20  X  10  =  0.073  henry. 
wL  —  0.073  ^  X57  =  11-26  ohms. 

R  =  0.3258  X  20  X  2  =  13.03  ohms. 

r  500,000 

1  1  =  -  ^—      —  ——  =  23.1  amperes. 
25,000  X  0.866 


I  =  ^(23.1  X  Q.866)2  +  (23.1  X  0.5  -  o.64)2  =  22'8  amPeres- 
Ed  =  22.8  v/(i3.o3)2  +  (n.26)2  =  392  volts. 
6  =  tan"1 0.864  =  40°  50'. 
1 1. 6  —  0.64 


20 


=0.548  =  28°  45 


=  12°  5'. 


"^(25,000  +  392  X  0.977)2  +'(392  X  0.209)"  =  25,338  volts. 


POWER  TRANSMISSION  AND   DISTRIBUTION  273 

If  greater  accuracy  is  required,  the  charging  current  may  be 
taken  as  that  due  to  the  mean  voltage  on  the  line,  and  a  second 
approximation  made.  This  is,  however,  seldom  necessary  as  it  is 
only  at  very  high  voltages,  or  over  long  distances  that  the  capaci- 
tance of  transmission  lines  becomes  very  marked,  and,  in  many 
cases,  it  may  be  disregarded  entirely  as  having  no  appreciable 
effect  on  the  voltage  regulation  of  the  line. 

Problem  —  Three  phase.  —  2000  kw.  are  to  be  transmitted  over  a 
distance  of  20  miles,  the  conductors  used  are  ooo  spaced  48  inches 
apart,  the  voltage  at  the  load  terminals  (line  to  line)  is  25,000,  the 
frequency  is  25,  and  the  power  factor  of  the  load  circuit  is  86.6  per 
cent.  Assume  the  capacitance  to  be  concentrated  at  the  ends  of 
the  line.  Find:  (a)  the  drop  in  the  line,  (b)  the  voltage  between 
the  generator  terminals. 

C  =  0.00155  X  5.28  X  20  =  0.1637  microfarad. 

L  I  4*_X  25  X  25,ooo_X  0.1637  X  io-*  = 

2  V3 

r       0.604  X  <;.28  X  20  X  io~3 

L  =  -  -  =  0.037  henry. 

coZ,  =  o.o  37  X  157  =  5.75  ohms  (per  line). 
R  =  0.3258  X  20  =  6.5  ohms  (per  line). 

2,000,000 

1  1  =  -  ~^=  =  53-5  amperes. 

25,000  X  0.866  X  V3 


I  =  ^(53-5  X  Q.866)2  +  (53-5  X  0.5  -  Q-37)2  =  53-3  amperes. 


Ed  =  53-3  v-S)2  +  (5-75)2  =  454  volts. 
6  =  tan"1  0.8847  =  4l0  3°' 
<£'  =  tan"1  0.569  =  29°  40'. 
6  -  0'  =  11°  50'. 


E  =  V  f^P2  +  454  X  o.978)2  +  (454  X  o.2o5)2 
\   V3  / 

=  14,895  volts  from  line  to  neutral 
—  25,768  volts  from  line  to  line. 

8.  Distributing  systems.  —  The  distributing  systems  in  common 
use  at  the  present  time  are:  (a)  series,  (b)  parallel,  (c)  series- 
parallel,  (d)  multiple  wire. 

(a)  The  series  system.  —  In  the  series  system  the  entire  current 
flows  successively  in  each  piece  of  apparatus,  the  voltage  varying 
as  the  load  changes,  and  the  current  remaining  approximately 
constant.  It  is  largely  used  in  alternating  current  street  lighting 


274  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

installations  in  connection  with  a  constant-current  transformer, 
either  with  or  without  a  mercury  arc  rectifier. 

(b)  The  parallel  system.  —  The  parallel  system  is  the  one  com- 
monly used  in  supplying  motors  and  incandescent  lamps.     The 
current  divides  before  reaching  the  lamps  or  the  motors,  so  that 
the  current  flowing  in  one  piece  of  apparatus  may  be  greater  or 
less  than  that  flowing  in  another  piece. 

(c)  Series-parallel  system.  —  It  is   undesirable   to   increase   the 
voltage  of  a  distributing  system  above  that  required  to  operate 
about  one  hundred  arc  lamps  connected  in  series.     Therefore,  in 
large  installations,  the  lamps  are  connected  into  series  groups,  and 
these  groups  are  connected  in  parallel,  each  group,  if  the  supply  is 
from  alternating-current  mains,  being  supplied  through  a  constant- 
current  transformer. 

(d)  Multiple-wire  systems.  —  When  the  power  in  a  system  is  con- 
stant, the  current  decreases  as  the  voltage  increases.     If  the  line 

__^ rrrrr    ^oss  ^^  *s  constant>  tne  resistance 

WIT     of   the  conductors  is   four  times  as 

itra  I  \\\\\ 

fance -I  11     great  when  the  voltage  of  the  system 
I  Converfer  -HI    is  doubled,  e.g.,  the  cross  section  (or 

FIG.   243-    The   Rotary   Converter    weight)    of    wjre    used    in    a    22O-volt 
as  a  Three- wire  Balancer.  .  . 

system  is  only  one-quarter  that  used 

in  a  no-volt  system,  the  line  loss  and  the  total  power  delivered  be- 
ing the  same  in  each  case.  This  fact  has  led  to  the  development  of 
multiple- wire  systems  which  effect  a  large  saving  in  copper  without 
increasing  the  voltage  between  terminals  of  the  load  apparatus. 
The  only  one  of  these  multiple-wire  systems  in  extensive  use  is  that 
using  three  wires,  a  greater  number  of  wires  adding  undesirable 
complications  in  both  the  generating  and  the  distributing  system. 

The  simplest  application  of  the  three-wire  system  is  in  connection 
with  an  alternating-current  system  where  the  third  or  neutral 
wire  is  connected  to  the  middle  point  of  the  secondary  winding 
of  a  transformer,  and  the  load  is  connected  between  each  of  the 
main  wires  and  the  neutral.* 

The  application  of  the  three-wire  principle  to  continuous-current 
distribution  requires  the  use  of:  (i)  a  rotary  converter  and  induc- 
tance coils,  (2)  a  special  generator,  (3)  a  motor-generator  balanc- 
ing set. 

*  See  Chapter  12,  Section  2. 


POWER  TRANSMISSION  AND   DISTRIBUTION 


275 


A  rmature      , 

Winding    I    POLE 

Brush, 


•  Positive  Wire 


- Neutral  Wire 


n  —  Meaa  five  Wire 
Auxiliary  Winding 


FIG.  244.     Schematic  Diagram  of  Crocker- 
Wheeler  Three-wire  Generator. 


Field  Rheostat 


(1)  Rotary  converter.  —  If  the  rings  of  a  rotary  converter  are 
connected  through  suitable  inductance  coils,  as  indicated  in  Fig. 
243,  the  converter  armature  acts  as  a  balance,  and  the  voltage 
between  either  main  and  a  neutral  connected  to  the  junction  of  the 
inductance   coils    is    approx- 
imately   equal    to    one-half 

that     between     the     mains, 

whatever  may  be  the  relative 

value  of  the  currents  in  the  commutator. - 

mains. 

(2)  Three-wire    generators. 
—  Three-wire  generators  are 
essentially  rotary  converters, 

the  neutral  connection  to  the  armature  being  made  through  an 
auxiliary  winding  or  through  inductance  coils. 

The  armature  of  the  Crocker- Wheeler  three-wire  generator  carries 
three  auxiliary  windings  which  are  star-connected  to  the  main 

winding  and  to  a  single  collector  ring, 
as  shown  in  Fig.  244.  The  Triumph 
Electric  Company's  three-wire  gen- 
erator makes  use  of  star-connected 
external  inductance  coils. 

(3)  The  motor-generator  balancer.  — 
If  two  similar  shunt  dynamos  are 
mounted  on  the  same  shaft  and  con- 
nected as  indicated  in  Fig.  245,  a  three-wire  system  is  produced. 
The  action  of  a  balancer  set  is  as  follows:  As  long  as  the  load  is 
balanced,  i.e.,  equal  currents  flow  in  the  main  wires,  zero  current 
flows  in  the  neutral,  and  the  currents  in  the  armatures  of  the 
balancer  set  are  equal.  When  the  currents  in  the  mains  are  un- 
equal, their  difference  flows  in  the  neutral,  and  the  current  in  the 
armature  connected  between  the  neutral  and  the  main  carrying 
the  smaller  current  increases.  This  increased  armature  current 
causes  the  speed  of  the  armature  to  increase,  and  the  other 
dynamo  is  driven  as  a  generator.  The  speed  of  the  balancer  set 
increases  until  equilibrium  is  reestablished,  and  the  voltage  between 
each  main  and  the  neutral  is  automatically  maintained  at  a  value 
equal,  approximately,  to  one-half  that  between  the  mains.  Neg- 
lecting the  losses,  the  armatures  of  the  balancer  set  are  required 


FIG.  245.    Motor-generator 
Balancer  Set. 


276  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

to  carry  the  current  flowing  in  the  neutral  wire,  and  this  current 
divides  inversely  as  the  voltages  between  the  neutral  and  the  main 
wires. 

The  voltage  regulation  of  a  three-wire  system  is  improved  by 
the  use  of  compound  balancers,  the  series  field  windings  being  so 

connected  that  the  motor  is  differentially 
AAiiiiiii    wound  and  the  generator  cumulatively 


<° )  wound. 

9.  Voltage  regulation  of  parallel  sys- 
tems. —  For  the  satisfactory  operation 
of  incandescent  lamps  the  electromotive 
force  between  the  terminals  of  the  lamps 


I  }'  }  {*   r  I  {  {  {    must  t>e  maintained  at  an  approximately 
(c)'  constant  value.     It   is   not   difficult    to 

design  a  wiring  system  for  a  group  of 

FIG.  246.    Parallel  Distri-        ,  , ,     ,    .  ,  -f   •          11 

bution  lamps  that  is  operated  as  a  unit,  ^.e.,  all 

the  lamps  in  the  group  are  either  lighted 

or  not  lighted,  but  to  design  a  wiring  system  for  the  same  group 
of  lamps,  any  part  of  which  may  be  in  use  at  a  given  time,  may 
be  a  rather  complicated  problem. 

The  general  problem  is  to  determine  the  center  of  distribution 
(the  center  of  gravity)  of  the  group,  at  which  the  voltage  should  be 
maintained  constant  by  means  of  feeder  regulation,  and  to  propor- 
tion the  wiring  between  this  center  and  the  individual  lamps,  so 
that  the  variation  of  voltage  at  the  most  disadvantageously  sit- 
uated lamp  never  exceeds  the  allowable  limits. 

To  meet  the  requirements  of  a  distributed  load,  various  wiring 
schemes  have  been  devised,  their  object  being  to  minimize  the  dif- 
ference in  voltage  between  the  terminals  of  different  lamps  in  the 
group.  Some  of  these  schemes  are  indicated  in  Fig.  246. 

10.  General  wiring  formulae.  —  The  following  expressions  will 
be  found  useful  in  the  solution  of  wiring  problems: 

p 

I  =  =—         -  for  single-phase  or  continuous-current  circuits. 
E  X  cos  0 

(9) 
p 

I  =  — - for  two-phase  circuits.  (10) 

2  E  X  cos  0 
p 

I  =  —- for  three-phase  circuits.  (n) 

V 3  E  X  cos  0 


POWER  TRANSMISSION   AND   DISTRIBUTION  277 

21.6  X  D  X  /  X  k  , 

Ed  =  -        -        -  for   continuous,   single-phase  or    two- 

\^s  •  1VJL  • 

phase  circuits.  (12) 

18.7  X  D  X  /  X  k  ,      ^ 

Ea  =  -  -  for  three-phase  circuits.  (13) 

when    C.M.  =  the  circular  mil  area  of  the  conductor  used, 
D  =  the  distance  of  transmission  in  feet, 
E  =  the  line  voltage  at  terminals  of  the  load, 
Ed  =  the  volts  lost  in  the  line. 
/  =  the  line  current, 

k  =  the  impedance  factor  /imPedanceN)  .     (Tables  XV  or 

\  resistance  / 

XVI.) 

Cos  0  =  the  power  factor  of  the  load  circuit, 

P  =  the  total  power  (watts)  delivered  to  the  load  circuit. 

Example.  —  30  k.w.  are  to  be  transmitted  over  a  distance  of  500 
feet,  the  voltage  at  the  terminals  of  the  load  is  440,  and  the  fre- 
quency is  60,  single-phase.  Find  the  drop  in  the  line  when  No.  4 
wires  spaced  18  inches  apart  are  used,  and  the  power  factor  is  0.85. 

—    *-       -  =  80.2  amperes. 


440  X  0.85 

,-,  ,       21.6  X  500  X  80.2  X  1.  12 

Ea  '  =  -  -  =  23.2  volts. 


Example.  —  100  k.w.  are  to  be  transmitted  a  distance  of  300  feet 
at  a  voltage  of  440,  a  frequency  of  60,  a  power  factor  of  0.85,  three- 
phase.  Find  the  size  of  wire  required,  the  line  drop  not  to  exceed 
4  per  cent. 

T  100,000 


V3  X  440  X  0.85 


=  155  amperes. 


From  Table  VIII  the  minimum  size  wire  allowable  is  No.  i  the 
area  of  which  is  83,690  c.m. 
Substituting  in  equation  (13) 

Ed  _  18.7  x  390x155x1.37  .       volts> 
83,690 

which  is  within  the  allowable  limits  and  should  be  used. 


278 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


FIG.  247.    Mershon's  Diagram. 

ii.   Mershon's    diagram.  —  Graphical   solution   of  alternating- 
current  distribution  problems  may  be  made  by  means  of  Mershon's 
diagram.*     Referring  to  Fig.  247  let 
the  radius  of  the  smallest  arc  represent  the  voltage  at  the  end  of 

the  feeder,  or  at  the  center  of  distribution; 
AB  represent  the  resistance  drop  in  the  line,  expressed  as  a  per- 
centage of  the  delivered  voltage; 

BC  represent  the  reactance  drop  in  the  line,  expressed  as  a  per- 
centage of  the  delivered  voltage. 

*  Originated  by  Ralph  D.  Mershon,  Past-President  of  the  American  Institute  of 
Electrical  Engineers. 


POWER  TRANSMISSION  AND   DISTRIBUTION 


279 


Lay  off  AB  beginning  at  a  point  A  where  the  arc  cuts  the  verti- 
cal representing  the  power  factor  of  the  load  circuit;  lay  off  BC 
perpendicular  to  AB,  as  indicated  in  the  diagram.  The  generator 
voltage  is  indicated  by  the  position  of  the  point  C  and  may  be  read 
directly  from  the  diagram.  The  power  factor  of  the  feeder  circuit 
is  determined  by  the  point  C  where  a  line  drawn  from  C  to  the 
origin  cuts  the  arc. 

12.  Feeder  regulation.  —  When  several  feeders  take  current 
from  the  same  bus  bars,  it  is  desirable  to  be  able  to  regulate, 
independently,  the  feeder  voltages.  Feeder  regulation  may  be 
accomplished  by  the  use  of:  (a)  boosters,  (b)  auto- transformers,  (c) 
induction  regulators.  Auto-transformers  and  induction  regulators 
are  applicable  to  alternating-current  circuits  only;  boosters  are 
used  on  continuous- current  circuits. 

(a)  Boosters.  —  A  line  booster  consists  of  a  series  generator  driven 
by  a  shunt  motor,  or  other  constant-speed  engine.  The  armature 


Switch 


\Auto-  trans  former    ,-Jj 


FIG.  248.    Diagram  of  Booster 
Connections. 


FIG.  249.      Diagram  of  Connections  for 
Auto-transformer  Feeder  Regulator. 


and  the  field  of  the  generator  are  connected  in  series  with  the  load, 
as  indicated  in  Fig.  248.  The  voltage  of  the  generator  is  propor- 
tional to  the  current  flowing  in  the  feeder  and  is  added  to  the 
bus-bar  voltage,  and  any  desired  increase  in  the  voltage  may  be 
obtained  by  properly  proportioning  the  booster  field. 

Because  of  the  addition  of  two  rotating  machines,  the  efficiency 
of  the  system  is  reduced  and  the  operating  complications  greatly 
increased. 

(b)  Auto-transformer.  —  If  an  auto-transformer  is  connected  as 
indicated  in  Fig.  249,  the  number  of  effective  turns  in  the  secondary 
depends  on  the  position  of  the  switch.  The  Stillwell  regulator 
operates  on  this  principle,  and  is  provided  with  a  reversing  switch  so 
that  the  secondary  voltage  may  either  "boost "  or  "buck  "  that  of  the 
bus  bars.  This  type  of  regulator  changes  the'voltage  by  definite  steps. 


280 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


(c)  Induction  regulators.  —  The  induction  regulator  is,  essentially, 
an  auto-transformer  in  which  the  angular  relations  of  the  primary 
and  secondary  coils  may  be  changed. 


(c)  Complete. 
FIG.  250.     Six-phase,  Motor-operated  Induction  Regulator.     General  Electric  Co. 

Structurally,  the  polyphase  induction  regulator  resembles  the 
polyphase  induction  motor  with  a  wound  rotor.  The  primary 
windings,  equal  in  number  to  the  number  of  phases  in  the  system 


POWER  TRANSMISSION  AND   DISTRIBUTION  281 

on  which  it  is  to  operate,  are  symmetrically  placed  in  slots  on  the 
surface  of  a  movable  laminated  iron  drum  (Fig.  25oa),  and  are 
connected  between  the  lines  of  a  polyphase  system.  The  secondary 
windings  are  symmetrically  placed  in  the  slots  of  a  stationary  core 
(Fig.  25ob),  and  are  connected  in  series  with  the  load  apparatus. 

The  primary  windings  set  up  a  rotating  flux  which  induces  a  con- 
stant electromotive  force  in  the  secondary  windings.  The  feeder 
voltage  is  the  vector  sum,  or  difference, 
of  the  bus-bar  voltage  and  the  voltage 
induced  in  the  secondary  winding  of  the 

regulator.  Since  the  phase  relation  of  FIG.  251.  Clock  Diagram  of 
the  bus-bar  voltage  and  the  induced  Polyphase  Induction  Regu- 
secondary  electromotive  force  depends  on  lator  Voltases- 
the  angular  position  of  the  movable  coils,  the  bus-bar  voltage  may 
be  raised  or  lowered  by  an  amount  equal  to  the  secondary  voltage 
of  the  regulator.  Fig.  251. 

The  single-phase  induction  regulator  differs,  both  in  principle  and 
in  structure,  from  the  polyphase  regulator.  A  schematic  diagram 

of  the  single-phase  regulator  is  shown 

in  Fig.  252.  The  voltage  induced  in 
the  secondary  coil  is  proportional  to 
the  cosine  of  the  angle  0  between 
the  axis  of  the  primary  coil  and  that 
of  the  secondary  winding.  The  in- 
duced electromotive  force  is,  there- 
fore, .zero  when  the  coils  are  at  right 
FIG.  252.  Schematic  Diagram  of  angles.  Since  the  flux  set  up  by  the 
Smgk -phase  Induction  Regu-  primary  coil  passes  through  the  sec- 
ondary coil  in  one  direction  when  the 

angle  0  is  less  than  90  degrees,  and  in  the  opposite  direction 
when  the  angle  /3  is  greater  than  90  degrees,  the  induced  electro- 
motive force  is  in  phase  with,  or  in  phase  opposition  to,  the  bus-bar 
voltage  as  the  angle  j3  is  greater  or  less  than  90  degrees. 

The  short-circuited  winding  which  is  placed  at  right  angles  to  the 
primary  coil  reduces  the  reactance  of  the  secondary  winding,  and 
thus  improves  the  power  factor  of  the  feeder  circuit.  By  reas  >n  of 
their  angular  relations,  the  effect  of  the  short-circuited  coil  increases 
as  that  of  the  primary  winding  decreases. 

Feeder  regulators  are  operated  either  by  hand  or  by  motor. 


282  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

When  motor  operated,  the  regulation  may  be  made  automatic  by 
means  of  a  contact-making  voltmeter  which  causes  the  motor  circuit 
to  be  closed  when  the  voltage  becomes  either  too  high  or  too  low. 
13.  Regulating  effect  of  a  constantly-excited  synchronous  motor. 
-  It  was  shown  in  Chapter  9  that  the  power  factor  of  a  synchro- 
nous motor  depends  on  the  relative  values  of  the  applied  and  the 
counter-electromotive  force.  Let  the  field  excitation  of  a  synchro- 
nous motor  at  the  end  of  a  feeder  be  such  that  the  power  factor  of 
the  feeder  circuit  is  unity.  If  the  load  increases,  the  voltage  at  the 
terminals  of  the  motor  drops,  the  current  leads  the  electromotive 
force,  and  the  leading  current  tends  to  increase  the  voltage  to  its 
former  value;  if  the  load  decreases,  the  voltage  at  the  terminals  of 
the  motor  increases,  the  current  lags  behind  the  electromotive  force, 
and  the  lagging  current  tends  to  decrease  the  voltage  to  its  former 
value.  The  tendency,  therefore,  of  a  constantly-excited  synchro- 
nous motor,  when  located  at  the  end  of  a  feeder,  is  to  maintain 
constant  voltage  at  the  center  of  distribution. 

14.  Voltmeter   compensation.  — 

Series  (differential) 

^^  _     When  feeder  regulators  are  used  it  is 


-shunt  winding         required  that  the  voltage  at  the  end  of 
the  feeder,  or  at  the  center  of  distribu- 

tion, be  indicated  by  an  instrument  at 
FIG.  ,53.    Voltmeter  Compensa-    ft     '        tf  For  'continuous.current 

tion  (C.  C.  Circuits). 

feeders,  it  is  only  necessary  that  a  prop- 

erly proportioned  series  winding  be  added  to  the  voltmeter.  Fig. 
253.  The  series  winding  is  so  connected  that  it  opposes  the  effect 
of  the  shunt  coil,  and  the  voltmeter  indication  is  reduced  by  an 
amount  equal  to  the  drop  in  the  line. 

Since  the  drop  in  an  alternating-current  feeder  is  not  a  simple 
ohmic  drop,  but  the  combined  effect  of  the  resistance  and  the  react- 
ance of  the  conductors,  the  voltage  at  the  end  of  an  alternating- 
current  feeder  is  indicated  by  a  voltmeter  only  when  the  voltage 
and  current  relations  in  the  instrument  circuit  are  identical  with 
those  in  the  feeder  circuit.  This  condition  is  effected  by  properly 
proportioning  the  resistance  and  the  reactance  of  the  secondary  cir- 
cuit of  a  series  transformer,  and  connecting  the  voltmeter  as  indi- 
cated in  Fig.  254.  Commercial  compensators  are  made  so  that  their 
resistance  and  reactance  may  be  varied,  thus  making  the  same  ap- 
paratus applicable  to  different  circuits. 


POWER  TRANSMISSION  AND   DISTRIBUTION 


283 


15.  Lightning  arresters.  —  The  effect  of  a  lightning  discharge 
on  electrical  apparatus  may  be  either  direct  or  indirect.  The  direct 
effect  is  the  destruction  of  the  insula- 
tion; the  indirect  effect  is  the  estab- 
lishment of  a  low-resistance  circuit 
which  may  be  maintained  by  the  nor- 
mal voltage  of  the  system.  A  satis- 
factory lightning  arrester  must,  then, 
divert  or  dissipate  the  energy  of  the 
discharge,  and  promptly  interrupt  any 
low-resistance  circuit  that  may  be  es- 


Feeder 

i. 

V 

Series 
Transforme 

Reactance 

*~z£*  SL 

Resistance 

tablished  for  the  purpose  of  diverting  FlG.  254.     schematic  Diagram 

or  dissipating  this  energy.  of  Connections  for  Voltmeter 

The  fundamental  operation  of  light-      Compensator  (A.  c.  Circuits). 
ning  arresters  will  be  explained  by  reference  to  Fig.  255  in  which 
G  is  an  air  gap  between  the  conductor  and  a  low-resistance  ground 
connection  and  A  is  an  inductive   coil  con- 
nected  in  series  with  the  line.     Under  normal 
operating  conditions  the  impedance  of  A  to 
ftie   flow   of  the   load   current,   either   con- 
tinuous  or  alternating,    is   small,   while  the 
FIG.   255.     Elementary  resistance  of  the  air  gap  G  is  so  large  that 

Lightning  Arrester.  . 

tne  leakage  is  negligible.  Since  the  impe- 
dance of  A  is  directly  proportional  to  the  frequency,  the  extremely 
high-frequency  lightning  (oscilla- 
tory) discharge  is  "  choked"  back, 
breaks  down  the  resistance  of  the 
air  gap  and  discharges  to  earth. 

It  is  a  well-known  fact  that 
an  electric  arc,  when  once  estab- 
lished, is  maintained  by  a  much 
smaller  voltage  than  is  required  to 
establish  it.  Consequently  the 
arc  established  by  the  lightning 
discharge  may  be  maintained  by 
the  normal  voltage  of  the  system  FlG*  256* 
unless  means  are  taken  to  suppress 
it.  This  is  done  by:  (a)  the  use  of  "  non-arcing"  metals,  (6)  the 
horn  gap. 


Westinghouse  Choke  Coil 
(Air  Cooled). 


284 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


(a)  Non-arcing  metal  arresters.  —  When  the  air-gap  terminals  are 
made  of  zinc  or  cadmium  or  of  their  alloys,  it  is  found  that  the  arc 
is  not  maintained  after  the  passage  of  the  lightning  discharge. 
Arresters  are,  therefore,  made  of  short  cylinders 
of  brass  separated  by  a  small  air  gap. 

Arresters  having  part  of  the  gaps  shunted  by 
resistance  have  come  into  extensive  use.  Their 
construction  is  based  on  the  theory  that,  for  a 
given  voltage,  high-frequency  discharges  require  a 
larger  number  of  gaps  than  do  discharges  of  low 
frequency.  Fig.  257  shows  a  2300- volt  General 
Electric  arrester  having  a  high  resistance  and  a 
low-resistance  shunt.  Low-frequency  discharges 
pass  through  the  high  resistance  (long  carbon  rod) 
and  two  air  gaps;  medium  frequency  discharges 
through  the  low  resistance  (short  carbon  rod)  and 
FIG.  257.  G.  E.  four  air  gaps;  while  high-frequency  discharges 

Lightning  Arrester.     ^  through  ^  ^^  ^^  Q£  ^  ^ 

Westinghouse  practice  differs  from  the  above  in  that  there  is  a 
shunt  and  a  series  resistance,  as  indicated  in  Fig.  258. 


'.Shunt 

•  Resistance 


;  Series  Resistance 
Ground 


Horn\  f 

ftp  V 


Line 


Ground 


FIG.  258.     Schematic  Diagram  of  West- 
inghouse Lightning  Arrester. 


FIG.  259.     Schematic  Diagram  of  Horn 
Gap  Arrester. 


;  (b)  Horn  gap  arrester.  —  The  horn  gap  arrester  has  the  appear- 
ance shown  in  Fig.  259,  and  is  connected  to  ground  in  series  with 
resistance.  The  lightning  discharge  breaks  down  the  air  gap  and 
forms  an  arc  between  the  horns.  Air  currents  formed  by  the  heat 
of  the  arc  carry  the  arc  itself  upward,  increasing  the  length  of  the 
arc  until  it  can  no  longer  be  maintained  by  the  normal  voltage  of 
the  system. 

During  the  past  few  years  there  has  been  developed  an  electrolytic 
cell  (Fig.  260),  which  replaces  the  resistance  used  in  connection  with 


POWER  TRANSMISSION   AND   DISTRIBUTION 


285 


the  horn  gap.  This  cell  consists  of  a  series  of  cone-shaped  aluminum 
elements,  the  spaces  between  which  are  partly  filled  with  electrolyte, 
and  the  whole  is  immersed  in  oil.  The  purpose  of  the  oil  is  to  in- 
crease the  insulation,  prevent  evaporation  of  the  electrolyte,  and 
dissipate  the  heat  liberated  during  discharge. 

The  action  of  the  cell  is  valve-like  in  that,  at  a  definite  critical 
voltage,  the  resistance  breaks  down,  and  a  small  increase  in  voltage 
above  this  critical  value  causes  a  large  current  to  flow.  When  the 
voltage  drops  below  the  critical  value,  which  is  about  40  per  cent 

above  the  normal  voltage  of  the 
system,  the  high-resistance  prop- 
erty of  the  cell  is  restored  and  the 
horn  gap  promptly  quenches  the  arc 
formed  by  the  discharge. 


Choke  Coil 


Ground 


FIG.  260.     Cross-section  of  General 
Electric  Electrolytic  Cell. 


FIG.  261.  An  Approved  Connection  for 
Three  -  phase  Electrolytic  Lightning 
Arrester. 


The  high  resistance  of  the  aluminum  cell  is  due  to  a  film  which  is 
deposited  on  the  aluminum  elements  when  a  current  flows  through 
the  cell.  Since  this  film  tends  to  dissolve,  the  cell  is  maintained  in 
operating  condition  by  periodically  bridging  the  horn  gaps  and 
allowing  a  momentary  discharge  through  the  cell. 

The  electrolytic  cell  is,  by  construction,  a  condenser  in  which,  for 
a  constant  applied  voltage,  the  current  is  directly  proportional  to 
the  frequency  (7  =  2  irfCE).  A  high-frequency  lightning  discharge 
is,  therefore,  diverted  through  the  cell  to  the  ground.  The  air  gap 
between  the  horns  prevents  the  small  leakage  current  that  would 
flow  if  the  cell  were  connected  directly  to  the  line. 


286 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


TABLE  VI 
DIMENSIONS,  WEIGHTS  AND  RESISTANCES  OF  COPPER  WIRES 

Diameter  =  — when  n  =  gauge  number. 

circular  mils 


Weight  (pounds)  per  1000  feet 
Resistance  (ohms)  per  1000  feet 


33° 

iQ.354 
circular  mils 


No. 
A.W.G. 
(B.  &  S.) 

Diameter, 
inches 

Area, 
C.M. 

Weight-pounds 

Resistance,  20°  C. 

looo  feet 

MUe 

1000  feet 

Mile 

oooo 

0.460 

211,600 

640.5 

338i 

o  .  04893 

0.2583 

ooo 

0.407 

167,800 

508.0 

2682 

0.06170 

0.3258 

oo 

0.365 

133,100 

402.8 

2127 

0.07780 

0.4108 

o 

0.325 

105,500 

319.5 

1687 

0.09811 

0.5180 

I 

0.289 

83,690 

253-3 

1337 

0.1237 

0.6531 

2 

0.258 

66,370 

200.9 

1062 

0.1560 

0.8237 

3 

0.229 

52,630 

159-3 

841.1 

0.1967 

1.0386 

4 

0.204 

41,740 

126.4 

667.4 

o  .  2480 

I.3094 

5 

0.182 

33.100 

IOO.2 

529-0 

0.3126 

1.6516 

6 

o.  162 

26,250 

79.46 

4I9-5 

o  .  3944 

2.0824 

7 

0.144 

20,820 

63.02 

332.7 

0-4973 

2-6257 

8 

0.129 

16,510 

49.98 

263.9 

0.6271 

3-3III 

9 

o.  114 

13,090 

39.63 

209.2 

o  .  7908 

4-1754 

10 

O.  IO2 

10,380 

31-43 

166.0 

0.9972 

5-2652 

ii 

0.091 

8,234 

24-93 

131.6 

1-257 

6.637 

12 

O.oSl 

6.530 

19.77 

104.4 

1.586 

8-374 

Resistance  of  aluminum  wire  =  160  per  cent  of  copper  wire. 
Weight  of  aluminum  wire  =  30  per  cent  of  copper  wire. 


POWER  TRANSMISSION  AND   DISTRIBUTION 


287 


TABLE   VII 

CURRENT  CARRYING-CAPACITY  OF  COPPER  WIRES 
National  Electric  Code 


A.W.G., 
(B.&S.) 

Sectional  area, 
circular  mils 

Rubber  insula- 
tion, amperes 

Other  insula- 
tion, amperes 

18 

1,624 

3 

5 

16 

2,583 

6 

8 

14 

4,107 

12 

16 

12 

6,530 

17 

23 

IO 

10,380 

24 

32 

8 

16,510 

33 

46 

6 

26,250 

46 

65 

5 

33-100 

54 

77 

4 

41,740 

65 

92 

3 

52,630 

76 

no 

2 

66,370 

QO 

131 

I 

83,690 

107 

156 

o 

105,500 

127 

i85 

oo 

133,100 

ISO 

220 

qco 

167,800 

177 

262 

oooo 

211,600 

2IO 

312 

200,000 

2OO 

3OO 

300,000 

27O 

OW 

4OO 

400,000 

0/v 

330 

*TWV 

i?OO 

500,000 

oow 
3QO 

O^*"7 

coo 

600,000 

ov 
450 

ovw 
680 

700,000 

*TO 

coo 

760 

800,000 

Ovv 

eeo 

840 

000,000 

00 

600 

W*f.W 

O2O 



1,000,000 

650 

y«w 

IOOO 

Current-carrying  capacity  of  aluminum  wire  =  75  per  cent  of  copper  wire. 


288 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


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290 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


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296 


ESSENTIALS   OF  ELECTRICAL  ENGINEERING 


TABLE  XVI 
LINE  VOLTAGE  AND  DISTANCE  BETWEEN  CONDUCTORS 


Length  of 

Distance  between 

Line  voltage 

transmission  line, 

wires, 

miles 

inches 

2,300 

i-     4 

12-   24 

6,600 

4-     8 

30-  36 

13,200 

8-  12 

36-  42 

22,OOO 

12-  18 

48-  60 

44,000 

1  8-  30 

60-   72 

66,000 

30-  50 

72-   84 

88.000 

50-  80 

96-108 

110,000 

80-125 

IO8-I2O 

While  the  above  voltages  have  become  standard  for  transmission  lines,  the 
distance  of  transmission  and  the  distance  between  wires  for  a  given  voltage 
vary  over  wide  limits,  and  the  above  table  is  intended  to  give  only  an  approx- 
imation to  average  values. 


TABLE  XVII 

Relative  weight  of  copper  required  for  different  systems  of  electrical  power 
distribution,  the  total  power,  the  distance  of  transmission,  the  line  loss  and 
the  voltage  at  the  load  terminals  remaining  constant. 


No.  of 

wires 

System 

Relative 
copper 

Load  connected 
between 

J 

Continuous-current  or   single-phase  alter- 

V IOO 

Line  wires 

2  1 

nating  current 

1 

J 

Three-wire    continuous-current    or    single- 

1     „    -   } 

Line  and 

6) 

phase  alternating  current,  neutral  full  size 
Three-wire    continuous-current    or    single- 

f    37-5  j 

neutral 
Line  and 

phase  alternating  current,  neutral  half  size 
Three  phase                     .          

75.0 

neutral 
Any  two  lines 

4 

Three  phase  with  neutral  .full  size  

33,  ) 

Line  and 
neutral 

4 

Three  phase  with  neutral  half  size  

29.2     j 

Line  and 
neutral 

Two  phase                                                

IOO.O 

Phase  wires 

3 

Two  phase  with  common  return  

72.9  1 

Line  and  com- 
mon return 

( 

5 

Two  phase  with  neutral  full  size  

31-25    { 

neutral 

POWER  TRANSMISSION  AND   DISTRIBUTION  297 

CHAPTER  XVIII  —  PROBLEMS 

1.  The  conductors  of  a  ioo-mile,*  3-phase,  25-cycle  transmission  system  are 
No.  o  copper  wire  spaced  10  feet.    The  voltage  at  the  load  end  (between  lines) 
is  104,000  when  10,000  kw.  are  being  delivered  to  a  receiving  circuit,  the  power 
factor  of  which  is  0.85.     Calculate  the  voltage  at  the  line  terminals  of  the 
step-up  transformers  supplying  the  transmission  line,  assuming  that  one-half 
the  capacitance  of  the  line  is  concentrated  at  each  end  of  the  line. 

2.  Same  as  Problem  i  except  the  capacitance  is  assumed  to  be  concentrated 
mid-way  between  the  generator  and  the  load. 

3.  Same  as  Problem  i  except  the  capacitance  is  assumed  to  be  concentrated 
as  follows:  one-sixth  at  the  generator,  one-sixth  at  the  load  and  two-thirds  mid- 
way between  the  generator  and  the  load. 

4.  Same  as  Problem  i  except  the  capacitance  of  the  line  is  neglected. 

5.  The  conductors  of  a  40-mile,  3-phase,  6o-cycle  transmission  line  are  £-inch 
aluminum f  wire  spaced  7  feet.    The  phase  voltage  at  the  load  is  38, 500  when  200 
amperes  flow  in  each  line.    The  power  factor  of  the  load  circuit  is  0.85.     Find: 
(a)  the  load  current,  (b)  the  voltage  at  the  generator  end  of  the  line,  assuming 
the  capacitance  to  be  concentrated  as  in  Problem  i. 

6.  Same  as  Problem  5  except  the  capacitance  is  neglected. 

7.  The  allowable  voltage  drop  in  a  single-phase,  6o-cycle  feeder  is  3  per  cent, 
and  the  distance  to  the  center  of  distribution  is  2000  feet.    Determine  the 
size  wire  required  when  the  voltage  at  the  load  is  2300,  the  power  delivered  is 
500  kw.,  and  the  power  factor  of  the  load  circuit  is  0.82.    Assume  spacing  of 
wires. 

8.  The  voltage  at  the  center  of  distribution  of  a  single-phase,  6o-cycle  feeder 
is  automatically  maintained  at  6600.    The  length  of  the  feeder  is  2  miles ;  No.  4 
copper  wires  are  used  and  spaced  3  feet.    The  load  on  the  feeder  is  400  kw.  and 
the  power  factor  of  the  load  circuit  is  0.75.    Determine:  (a)  the  volts  drop  in  the 
line,  (b)  the  voltage  at  the  generator. 

9.  A  5oo-horse-power,  25-cycle,  3-phase  induction  motor  is  300  feet  from  the 
generator  supplying  it  with  current.    Efficiency  of  motor  =  90  per  cent.     Power 
factor  of  motor  =  0.85.    Determine  the  size  of  wire  required  so  that  the  voltage 
drop  between  no  load  and  full  load  shall  not  exceed  10  per  cent. 

10.  100  alternating-current  series  arcl  amps  are  operated  on  a  certain  cir- 
cuit of  No.  6  copper  wire  (diameter  =  0.162  inch).    The  first  and  the  last 
lamp  in  the  circuit  are  each  300  feet  from  the  power  house,  and  the  lamps 
are  spaced  100  feet  apart.    Each  lamp  takes  40  volts  and  6.6  amperes.    The 
lamps  operate  at  a  power  factor  of  0.85.    Find:   (a)  the  drop  due  to  the 
resistance  of  the  line,  (b}  the  power  lost  in  the  line,  (c)  the  power  input  to  the 
system,  (d)  the  power  factor  of  the  system. 

*  Because  of  the  "sag"  and  the  contraction  and  expansion  due  to  changes  in  tem- 
perature, the  length  of  wire  required  is  approximately  10  per  cent  greater  than  the 
distance. 

t  Aluminum  conductors  are  always  stranded,  and  their  nominal  diameter  is  the 
solid  equivalent  of  the  actual  diameter. 


298  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

11.  The  input  to  an  induction  motor  installation  is  400  kw.  at  a  power  factor 
of  0.80.    Find  the  rating  of  a  synchronous  condenser,  the  power  factor  of  which, 
is  o.io,  to  make  the  power  factor  of  the  feeder  circuit  unity. 

12.  Find  the  power  factor  of  the  feeder  circuit  in  Problem  n  when  the  in* 
duction  motor  load  is  increased  to  500  kw. 


CHAPTER  XIX 
THE   STORAGE  BATTERY* 

BY  means  of  a  storage  battery,  energy  may  be  stored  and  made 
available  at  some  future  time.  Structurally  the  storage  battery 
consists  of  a  set  of  positive  plates,  a  set  of  negative  plates,  and  a 
chemical  solution  (electrolyte)  in  which  the  plates  are  immersed. 


Positive  Plate  Negative  Plate 

FIG.  262.     "Ironclad-Exide"  Storage  Battery.    The  Electric  Storage  Battery 

Company. 

The  internal  actions  of  a  storage  battery  are  chemical  and  not 
mechanical  as  the  name  might  imply.  When  an  electric  current 
passes  through  the  electrolyte  from  the  positive  to  the  negative  plates, 
it  produces  a  chemical  reaction  and  a  structural  change  in  the 
"active  materiar'  of  the  plates.  When  the  positive  and  the  nega- 
tive plates  are  connected  by  means  of  a  metallic  wire  or  other 
electrical  conductor,  an  inverse  chemical  action  takes  place,  the 
plates  are  restored  to  their  original  condition,  and  an  electric  cur- 
rent flows  in  the  wire  from  the  positive  to  the  negative  plates. 

*  For  details  of  storage  battery  construction,  operation  and  maintenance,  the 
student  is  referred  to  "Secondary  Batteries"  by  E.  J.  Wade. 

299 


300 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


i.  Lead  batteries.  —  In  the  completely  discharged  condition, 
the  plates  of  lead  storage  batteries  are  lead  sulphate  (PbSO4),  and 
the  electrolyte  is  water.  When  a  unidirectional  current  passes 
through  this  cell  from  the  positive  to  the  negative  plates,  the 
chemical  reaction  produces  peroxide  of  lead  at  the  positive  plate, 


Negative  Group 


Wood  Separator 


Positive  Group 


Glass  Jar,  Oak  Sand  Tray  with 

Sand  Glass  Insulators 
FIG.  263.     Storage  Battery  Parts.     Gould  Storage  Battery  Co. 

" sponge"  lead  at  the  negative  plate,  and  converts  the  electrolyte 
into  sulphuric  acid.     This  action  may  be  represented  as  follows: 


Discharged 

Charging  current 

Charged 

Discharging  current 


Positive  plate 

PbSO4 


Electrolyte 

2H2O 


Pb02 


2  H2SO4 


Negative  plate 

PbSO4 
Pb 


While  other  reactions  may,  and  probably  do,  take  place  in  the  cell, 
the  scientific  world  generally  accepts  the  above  as  representing  the 
fundamental  reactions. 

The  plates  used  in  lead  storage  batteries  are  of  two  general  types, 
differentiated  by  the  process  of  their  manufacture :  (a)  Plante  type 


THE  STORAGE  BATTERY  301 

in  which  the  active  material  of  the  plates  is  formed  by  chemical 
processes,  (b)  Faure  type  in  which  the  active  material  is  prepared 
and  applied  mechanically  (pasted)  to  a  supporting  frame  or  grid 
through  which  the  electric  current  flows.  The  plates  in  commer- 
cial batteries  are  often  a  combination  of  the  above  types. 

The  open-circuit  voltage  of  a  fully  charged  lead  cell  is  about  2.2. 
Because  of  internal  resistance,  which  is  not  constant,  this  voltage 
is  not  realized  when  the  cell  is  discharging,  and  the  voltage  gradually 


FIG.  264.     Edison  Storage  Battery. 

decreases  as  the  cell  discharges,  as  indicated  in  Fig.  265.  The  prac- 
tical limit  of  discharge  is  reached  when  the  voltage  is  1.75,  and  the 
current  is  that  at  which  the  cell  is 'rated.  Further  discharge  causes 
an  excess  of  lead  sulphate  to  form  on  the  plates,  increase  the  in- 
ternal resistance*  of  the  cell,  and  sets  up  stresses f  which  cause 
the  plates  to  crack  and  fall  away  from  the  supporting  grid. 

2.  The  Edison  battery.  —  The  Edison  storage  battery  consists  of 
a  positive  plate  of  nickel  hydrate  and  metallic  nickel,  and  a  negative 
plate  of  the  oxides  of  iron  and  mercury,  immersed  in  a  20  per  cent 
solution  of  caustic  potash.  The  purpose  of  the  metallic  nickel  and 
the  mercury  oxide  is  to  increase  the  conductivity  of  the  active  mate- 
rials of  the  plates.  The  electrolyte  undergoes  no  chemical  change 
during  either  charge  or  discharge,  acting  simply  as  a  medium  through 
which  oxygen  is  transferred  from  one  plate  to  the  other.  The  charg- 
ing current  reduces  the  oxide  of  iron  (negative  plate)  to  a  mass  of 
"sponge"  iron,  and  produces  peroxide  of  nickel  at  the  positive  plate. 

*  Lead  sulphate  is  a  non-conductor  of  electricity, 
t  Because  of  the  change  in  volume. 


302 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


As  shown  in  Fig.  265,  the  voltage  of  the  Edison  cell  is  less  than 
that  of  the  lead  cell,  and  decreases  more  rapidly  as  the  cell  dis- 
charges. The  cell  is  not  injured  by  abnormal  or  complete  dis- 
charge. Its  capacity  per  unit  weight  is  greater  than  that  of  the 


I  234  5676 

TIME  IM  HOURS 

FIG.  265.     Charge  and  Discharge  Curves  for  Storage  Cells. 


lead  cell,  but  its  energy  efficiency  is  only  60%  as  compared  with 
80%  in  the  lead  cell. 

3.  Storage  battery  ratings.  — Storage  batteries  are  rated  in  am- 
pere-hours, and  the  normal  discharge  period  of  the  lead  is  eight 
hours,  i.e.,  when  the  discharge  current  is  constant  and  normal,  the 
voltage  of  the  lead  cell  drops  to  1.75  in  eight  hours.  The  capacity 
of  the  cell  is  decreased  when  the  rate  of  discharge  is  increased,  as 
shown  by  Table  XVIII.  As  shown  in  Fig.  265,  the  normal  period 
for  the  charge  or  discharge  of  an  Edison  cell  is  five  hours. 

TABLE  XVIII 


Time  of  dis- 

Relative 

Time  of  dis- 

Relative 

charge,  hours 

capacity 

charge,  hours 

capacity 

8 

100 

4 

83 

7 

97 

3 

75 

6 

93 

2 

65 

5 

89 

I 

50 

4.   Applications  of  the  storage  battery.  —  The  following  are  a  few 
of  the  more  important  applications  of  the  storage  battery: 


THE   STORAGE   BATTERY 


303 


(a)  To  furnish  energy  during  periods  of  light  load  when  the 
generating  apparatus  is  shut  down. 

(b)  To  aid  in  carrying  peak  loads  which  would  otherwise  overload 
the  generating  apparatus. 

(c)  To  maintain  a  more  nearly  constant  voltage  with  varying  load. 

(d)  To  maintain  a  more  nearly  uniform  load  on  the  generating 
apparatus,  the  battery  charging  during  periods  of  light  load  and  dis- 
charging during  periods  of  heavy  load. 

From  the  nature  of  its  internal  actions,  it  is  evident  that  a  storage 
battery  cannot  be  charged  from  an  alternating-current  system 
without  using  a  rectifying  device,  such  as  a  rotary  converter  or  a 
motor-generator. 

5.  Storage-battery  connections. — In  Fig.  266,  the  battery  is  con- 
nected in  parallel  with  the  load,  and  is  preferably  placed  at  the  end 
of  a  feeder.  As  the  load  on  the  feeder  increases,  the  line  drop  re- 
duces the  voltage  at  the  terminals  of  the  battery,  and  the  battery 
discharges;  as  the  load  on  the  feeder  decreases,  the  voltage  at  the 
terminals  of  the  battery  increases,  and  the  battery  charges. 


Battery 


T 


FIG.  266.     "Floating"  Storage  Battery. 


FIG.  267.    Storage  Battery  with  End- 
cell  Regulation. 


In  Fig.  267,  end-cell  switches  are  provided  so  that  either  the  charg- 
ing or  the  discharging  voltage  may  be  regulated.  With  this  con- 
nection the  load  voltage  is  independent  of  the  generator  voltage. 

In  Fig.  268,  the  generator  of  the  motor-generator  set  is  differ- 
entially wound  and  has  its  armature  connected  in  series  with  the 
battery.  The  field  windings  are 
so  proportioned  that  the  voltage 
of  the  generator  is  zero  when 
the  load  on  the  system  is  at  its 
average  value,  and  the  battery 

neither   charges   nor    discharges.    FlG-  268-     Storage  Battery  with  Differen- 

If  the  load  increases,  the  differ-  tial  Booster' 

ential  series  field  reduces  the  generator  voltage,  and  the  battery 
discharges;  if  the  load  decreases,  the  generator  voltage  increases 
and  the  battery  charges. 


Generator 


304  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

6.  Limitations  of  the  storage  battery.  —  The  commercial  use  of 
the  storage  battery  is  limited  because  of: 

(a)  Its  high  first  cost  and  rapid  depreciation. 

(b)  The  additional  mechanical  complications  introduced  by  reason 
of  the  regulating  apparatus  required. 

(c)  The  cost  of  maintenance. 

7.  Care  of   Storage  Batteries.  —  Explicit  instructions  for  the 
care  of  storage  batteries  cannot  be  given  because  service  condi- 
tions, as  well  as  the  ideas  of  different  manufacturers,  differ  very 
radically.     It  is  reasonable  to  assume  that  a  manufacturer  knows 
the  conditions  under  which  his  apparatus  will  give  the  best  results, 
and  storage  batteries  should  be  operated  in  strict  accordance  with 
the  instructions  of  the  manufacturer. 


APPENDIX  A 


HARMONIC    QUANTITIES 

1.  Definition.  —  A  harmonic  quantity  is   a   quantity,  the   instantaneous 
value  of  which  is  proportional  to  the  sine  or  the  cosine  of  a  uniformly  increas- 
ing angle,  and  the  constant  of  the  equation  represents  the  maximum  value  of 
the  quantity.     Common  examples  of  harmonic  quantities  are:   (a)  the  velocity 
of  a  pendulum,  (6)  the  vibration  of  a  tuning  fork,  (c)  the  velocity  of  the  cross- 
head    of    a   reciprocating    engine* 

when  the  flywheel  rotates  at  a  uni- 
form angular  velocity. 

Let  OA  (Fig.  269)  represent  the 
maximum  value  of  any  harmonically 
varying  quantity.  The  value  of  the 
quantity  at  any  given  instant  or  FlG  26g  Harmonic  or  Sine  Curve, 
position  is  proportional  to  the  pro- 
jection of  OA  on  the  vertical  axis,  i.e.,  to  the  sine  of  the  angle  #. 

Ofl  =  (X4sin$.  (i) 

If  the  angular  velocity  of  the  vector  OA  is  expressed  in  circular  measure 
(radians), 

<t>  =  ut  (2) 

and  Oa  =  OA  sin  co/,  (3) 

when  a)  =  the  angular  velocity  (radians  per  unit  of  time)  at  which  the  vector 

rotates, 

/  =  the  time  during  which  the  vector  has  rotated,  zero  time  coinciding 
with  zero  value  of  the  angle  <f>. 

2.  Rectangular  representation.  —  A  sine  wave  may  be  plotted  to  rectangu- 
lar coordinates  as  shown  in  Fig.  269,  the  curve  there  plotted  showing  a  succes- 
sion of  values  for  one  complete  revolution  (360  degrees).     A  representation  of 

greater  values  of  ut  would  be  a  repetition  of  those  values 
already  plotted. 

3.   Polar  representation.  —  It  is  usually  unnecessary  to 
draw  rectangular  representations  of  harmonic  quantities, 
the  magnitudes  of  the  quantities  and  their  relative  posi- 
FIG.  270.     Vector   tions  being  sufficient>     Such  a  diagram  (Fig.  270)  is  called 


or     Clock 
gram 


Dia- 


a  vector  or  clock  diagram,  and  the  vector  rotates  in  a 
counter-clockwise  direction. 
4.    Combination  of  harmonic  quantities.  —  Two  or  more  harmonic  quantities, 
having  the  same  angular  velocities,  may  be  replaced  by  a  single  harmonic 

*  This  velocity  is  only  approximately  harmonic  when  the  connecting  rod  is  of  finite 
length. 

305 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


quantity,  the  vector  of  which  is  the  geometric  sum  of  the  vectors  of  the  two 
quantities. 

Let  OA  (Fig.  271)  represent  the  maximum  value  of  one  harmonic  quantity 
and  OB  the  maximum  value  of  another  harmonic  quantity  having  the  same 
angular  velocity.  Then 

Oa  =  OA  sin  »/  (4) 

and  Ob  =  OB  sin  (ut  —  <t>).  (5) 


FIG.  2yia.    Combination  of  Vectors.     FIG.  27  ib.    Combination  of  Sine  Waves. 

Combining  these  two  vectors  graphically,  the  vector  OC  is  obtained,  for  which 
the  mathematical  expression  is 


(7) 
(8) 


-  0).  (6) 

The  mathematical  proof  that  the  resultant  of  two  harmonic  quantities  is  a 
harmonic  quantity  having  the  same  angular  velocity  is  as  follows:  Let 
a  =  A  sin  w/, 
b  =  £sin(co/-0). 


Then  a  +  b  =  A  sin  at  +  B  sin  (wt  —  <f>). 

Expanding  equation  (9) 

a  +  b  =  A  sin  ut  +  B  sin  wt  cos  <£  —  B  cos  o>/  sin  <f> 

=  (A  +  B  cos  0)  sin  at  —  B  cos  co/  sin  <fr. 
But  ^4  +  B  cos  </>  =  C  cos  0 

and  5  sin  0  =  C  sin  0 

— !  =  tan"  =  T 

COS0  yl 

Substituting  in  equation  (n) 

a  +  b  =  \/(Az  +  B2  —  2  AB  cos  <£)  sin  (  o>/  — 

From  equation  (15)  the  maximum  value  of  the  resultant  is 
C  = 


(9) 

(10) 

(n) 
(12) 

d3) 
(14) 


V 
/J 


B*-2ABcos<j> 

and  the  angle  through  which  the  vector  has  rotated  is 

.      B  sin  <£ 

ft  =  tat  —  tan"1  .   ,    „          • 
4  4-  £  cos  <j> 


(16) 
(17) 


APPENDIX  A  307 

5.  Phase  difference.  —  The  phase  (time)  difference  between  two  harmonic 
quantities  having  the  same  angular  velocities,  is  the  time  required  for  the  vector 
of  either  to  pass  through  an  angle  equal  to  that  between  the  vectors  of  the  two 
quantities.    The  phasa  angle,  or  angle  of  phase  difference,  is  the  angle  AOB  in 
Fig.  27ia. 

6.  Resolution  of  harmonic  quantities.  —  If  two  harmonic  quantities  may  be 
combined  into  a  single  harmonic  quantity  having  the  same  angular  velocity,  it 
follows  that  any  harmonic  quantity  may  be  resolved  into  harmonic  components 
having  any  desired  phase  difference.    In  alternating-current  problems,  the 
resolution  of  harmonic  electromotive  forces  and  currents  into  components 
having   a   phase  difference  of  90  degrees,  is  a  common  and  a  useful  ex- 
pedient. 

7.  Rate  of  change  in  the  instantaneous  value  of  a  harmonic  quantity.  — 
An  inspection  of  a  sine  curve  such  as  that  shown  in  Fig.  269,  shows  that  the 
rate  at  which  its  instantaneous  value  changes  is  not  uniform,  and  that  the 
rate  of  change  is  greatest  when  the  quantity  is  passing  through  its  zero 
value. 

This  may  also  be  shown  by  differentiating  the  expression  for  the  instanta- 
neous value 

a  =  A  sin  ut  (18) 

with  respect  to  /.    The  rate  at  which  a  changes  is 

-~  =  uA  cos  ut.  (19) 

But  cos  o>/  is  maximum  when  sin  o>t  is  zero. 

8.  Average  value  of  the  sin  w/.  —  The  average  value  of  the  sin  wt  for  a  com- 
plete cycle  is  zero,  since  for  each  positive  value  there  is  an  equal  negative  value. 

For  any  half  cycle  beginning  with  o>/  =  o  or  ut  =  T,  all  the  values  are  either 
positive  or  negative  and  the  average  may  be  determined  by  adding  together 
the  values  given  in  Table  XIX,  and  dividing  by  the  number  of  values.  This 
gives 

av.  sin  w/  =  ±  0.637.  (2°) 

The  same  result  is  obtained  by  integrating  sin  w/  between  the  limits  <•>/  =  » 
and  co/  =  o,  and  dividing  by  *-. 

•j    ~w/  =  f 
av.  sinw/=-|          sinco/(f(aJ)  (21) 

7T  Jw<=0 


9.  Value  of  average  sin2  «/.  —  Let  the  ordinates  of  a  curve  be  equal  to  the 
squares  of  the  instantaneous  values  of  a  harmonic  quantity. 

y  =  <**  (23) 

=  A*  sin2  co/.  (24) 


308 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


TABLE  XIX 
NATURAL  SINES,  COSINES,  TANGENTS  AND  COTANGENTS 


A 

Sin 

Cos 

Tan 

Cot 

0 

O  .  OOOOO 

I  .  OOOOO 

O  .  OOOOO 

Infinity 

90 

I 

0.01745 

0.9998 

0.01745 

57.2900 

89 

2 

0.03490 

0.9994 

0.03492 

28.6363 

88 

3 

0.05234 

0.9986 

0.05241 

19.0811 

87 

4 

0.06976 

0.9976 

0.06993 

14.3007 

86 

5 

0.08716 

0.9962 

0.08749 

11.4301 

85 

6 

0.10453 

0-9945 

o.  10510 

9-5I44 

84 

7 

o.  12187 

0.9925 

o.  12278 

8  .  1443 

83 

8 

0.1392 

o  .  9903 

0.1405 

7-U54 

82 

9 

0.1564 

0.9877 

0.1584 

6.3138 

81 

10 

0.1736 

o  .  9848 

0.1763 

5-6713 

80 

ii 

o.  1908 

0.9816 

0.1944 

5  •  H46 

79 

12 

0.2079 

0.9781 

o.  2126 

4.7046 

78 

13 

0.2250 

0.9744 

0.2309 

4.3315 

77 

14 

s  0.2419 

0.9703 

0.2493 

4.0108 

76 

15 

0.2588 

0.9659 

0.2679 

3-7321 

75 

16 

0.2756 

0.9613 

0.2867 

3-4874 

74 

17 

0.2924 

0.9563 

0.3057 

3.2709 

73 

18 

0.3090 

0.9511 

0.3249 

3.0777 

72 

19 

0.3256 

0.9455 

o  .  3443 

2.9042 

7i 

20 

0.3420 

0.9397 

o  .  3640 

2-7475 

70 

21 

0.3584 

0.9336 

0.2839 

2.6051 

69 

22 

0.3746 

0.9272 

o  .  4040 

2-4751 

68 

23 

0.3907 

0.9205 

0.4245 

2-3559 

67 

24 

0.4067 

0.9135 

0.4452 

2  .  2460 

66 

25 

0.4226 

0.9063 

o  .  4663 

2  •  1445 

65 

26 

0.4384 

0.8988 

0.4877 

2.0503 

64 

27 

o  .  4540 

0.8910 

0.5095 

I  .9626 

63 

28 

0.4695 

0.8829 

0.5317 

I  .  8807 

62 

29 

o  .  4848 

0.8746 

0.5543 

I  .  8040 

61 

3° 

o  .  5000 

0.8660 

0.5774 

I.732I 

60 

3i 

0.5150 

0.8572 

o  .  6009 

1-6643 

59 

32 

0.5299 

o  .  8480 

0.6249 

I  .  6003 

58 

33 

o  .  5446 

0.8387 

o  .  6404 

1-5399 

57 

34 

0.5592 

0.8290 

0.6745 

1.4826 

56 

35 

0.5736 

0.8192 

o  .  7002 

i  .4281 

55 

36 

0.5878 

0.8090 

0.7265 

1.3764 

54 

37 

0.6018 

o  .  7986 

0.7536 

1.3270 

53 

38 

0.6157 

o  .  7880 

0.7813 

1.2799 

52 

39 

0.6293 

0.7771 

0.8098 

i  •  2349 

Si 

40 

0.6428 

o  .  7660 

0.8391 

1.1918 

50 

4i 

0.6561 

0.7547 

o  .  8693 

1.1504 

49 

42 

0.6691 

o.743i 

o  .  9004 

i  .1106 

48 

43 

0.6820 

0-73U 

0.9325 

1.0724 

47 

44 

0.6947 

0.7193 

0.9657 

1-0355 

46 

45 

0.7071 

0.7071 

I  .0000 

I.  0000 

45 

Cos 

Sin 

Cot 

Tan 

A 

For  a  more  complete  table  of  functions  see  any  standard  text  on  trigonometry. 


APPENDIX  A  309 

Integrating  equation  (24)  between  the  limits  co/  =  TT  and  co/  =  o,  and  dividing 
by  •*, 

av.  y  =  —  f"1    'sin2  co/  d  (tat)  (25) 


0 

-f  ;  (*) 

i.e.,  the  average  value  of  sin2  co/  equals  £. 

The  same  result  may  be  deduced  without  recourse  to  calculus.  From 
trigonometry 

sin2  w/  +  cos2  co/  =  i,  (27) 

and  av.  sin2  co/  -f  av.  cos2  co/  =  i.  (28) 

As  co/  varies  from  o  to  90  degrees,  the  sine  passes  through  all  values  from  o  to  i 
and  the  cosine  passes  through  all  values  from  i  to  o.    Therefore, 

av.  sin2  co/  =  av.  cos2co/  (29) 

and  av.  sin2  co/  =  i.  (30) 

10.  Frequency.  —  The  frequency  of  a  harmonic  quantity  is  the  number  of 
complete  revolutions  made  by  its  vector  per  unit  of  time.  During  one  cycle  a 
harmonic  quantity  passes  through  all  possible  instantaneous  values,  both  posi- 
tive and  negative,  i.e.,  starting  at  zero  the  quantity  increases  to  maximum, 
decreases  to  zero,  increases  to  maximum  in  the  opposite  direction  and  again 
decreases  to  zero. 

n.  Value  of  «.  —  Since  the  distance  passed  through  by  a  rotating  vector  is 
expressed  in  radians,  the  angular  velocity  (radians  per  second)  is  equal  to  the 
frequency  (number  of  revolutions  the  vector  makes  in  one  second)  multiplied 
by  2  TT. 

co  =  2*/.  (31) 

APPENDIX  A  —  PROBLEMS 

1.  Find  the  resultant  of  two  harmonic  quantities,  each  of  which  has  a  maxi- 
mum value  of  1000,  when  the  angle  between  their  vectors  is:  (a)  30  degrees,  (b) 
45  degrees,  (c)  60  degrees,  (d)  90  degrees,  (e)  120  degrees. 

2.  The  maximum  value  of  a  harmonic  quantity  is  600,  and  the  angle  between 
its  (two)  components  is  90  degrees.     Find  the  maximum  values  of  the  compo- 
nents when:  (a)  the  angle  between  the  quantity  and  one  component  is  30  degrees, 
(b)  the  maximum  values  of  the  components  are  equal,  (c)  the  maximum  value 
of  one  is  twice  the  maximum  value  of  the  other. 

3.  Find  the  angles  of  phase  difference  between  the  quantity  in  Problem  2 
and  its  components. 

4.  Find  the  angular  velocity  of  a  rotating  vector  when  the  frequency  is:  (a) 
20,  (b)  25,  (c)  30,  (d)  40,  (e)  50,  (/)  60,  (g)  100. 

5.  The  maximum  value  of  a  harmonic  quantity  is  100.     Find  the  rate  at 
which  the  quantity  is  changing  when  co/  is  equal  to:  (a)  o,  (b)  30  degrees,  (c)  45 
degrees,  (d)  60  degrees,  (e)  75  degrees,  (/)  90  degrees. 


310  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

6.  Find  the  average  value  of  a  harmonic  quantity,  the  maximum  value  of 
which  is:  (a)  40,  (6)  75,  (c)  100,  (d)  250,  (e)  800. 

7.  Find  the  average  square  of  a  harmonic  quantity,  the  maximum  value  of 
which  is:  (a)  40,  (b)  75,  (c)  100,  (d)  250,  (e)  800. 

8.  Find  the  frequency  when  «  is  equal  to:  (a)  125.6,  (b}  157,  (c)  188.4,  (d) 
251.2,  (e)  314,  (/)  376.8,  (g)  400. 


APPENDIX  B 
INDUCTANCE 

In  Chapter  i,  Section  n,  inductance  is  denned  as  the  proportionality  factor 
between  the  electromotive  force  set  up  in  a  circuit  by  reason  of  a  change  in  the 
value  of  the  current  flowing  in  the  circuit,  and  the  rate  at  which  the  current 
changes. 


From  Chapter  2,  Section  13 


Therefore,  L      =  (3) 

and  L  =  *>  (4) 

when  <f>  =  the  total  flux  linking  with  the  circuit  (if  the  conductor  has  more  than 
one  turn  or  loop,  <j>  is  the  product  of  the  flux  linking  with  one  turn 
and  the  number  of  turns), 

*  =  the  current  flowing  in  the  circuit,  in  c.g.s.  units, 
e  =  the  electromotive  force  of  self-induction,  in  c.g.s.  units, 
L  =  the  inductance  of  the  circuit,  in  c.g.s.  units. 

i.  Inductance  of  each  of  two  parallel  wires.  —  Let  A  and  B  be  two  parallel 
cylindrical  conductors  in  which  the 
currents  are  equal  but  flow  in  opposite 
directions.  The  current  in  each  conduc- 
tor,  if  uninfluenced  by  that  in  other  con- 
ductors, would  set  up  concentric  circles 
of  flux  around  the  axis  of  the  wire 
(Chapter  2,  Section  3).  The  mutual 
effect  of  the  currents  is  to  produce  the  FlG-  2?2-  Magnetic  Field  Between  Two 
flux  distribution  indicated  in  Fig.  272.  Current  <***&*  Conductors. 

The  inductance  of  the  conductors  is  due  to  the  flux  which  passes  between 
the  axes  of  the  wires,  the  flux  which  encircles  A  at  a  greater  distance  than  D 
being  neutralized  by  an  equal  and  opposite  flux  set  up  by  B.  The  total 
flux  encircling  the  axis  of  each  wire  may  be  divided  into  two  parts:  (a)  that 
in  the  body  of  the  wire,  (6)  that  in  the  insulating  material  between  the 
wires. 

3" 


312  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

(a)  The  flux  in  the  body  of  the  wire.  —  The  flux  in  any  elemental  zone  dx 
(Fig.  273)  within  the  wire  is  due  to  the  current  inside  of  zone  dx,  and  this 
current  is  ^2  * 


FIG.  ,73-  =      P'  (7) 

Therefore,  the  flux  in  zone  dx  is 

2  9ci 

<i>  =  —  -  dx  c.g.s.  units.  (8) 

Since  this  flux  surrounds  only  that  part  of  the  current  which  flows  in  the  area  of 
the  conductor  bounded  by  the  zone  dx,  it  is  equivalent  to 

*'-Jx^<fa    '  :•••;  (9) 

2  -v3'£ 

=  —  —  dx  c.g.s.  units,  (10) 

f  ',,•:-.. 

linking  with  the  entire  current  flowing  in  the  conductor.  The  total  flux  per 
centimeter  length  of  conductor  within  the  area  of  the  conductor  is  found  by 
integrating  equation  (10)  between  the  limits  x  =  r  and  x  =  o. 


Jf 


=  -c.g.s.  units.  (12) 

2 

(6)  The  flux  in  the  insulating  material  between  the  wires.  —  The  flux  passing 
between  the  conductors  is  due  to  the  current  flowing  in  the  wires. 


Therefore,  the  flux  in  zone  dx  (Fig.  274)  due  to  the  current  in  A  is 

4,"  =  —dx  c.g.s.  units,  (15) 

*  If  the  current  density  in  the  conductor  is  not  uniform  this  expression  becomes  an 
approximation  only. 

t  The  wires  are  here  assumed  to  be  copper  or  aluminum,  the  permeability  of  which 
is  unity.  If  the  conductors  are  of  magnetic  material  (iron)  B  =  pH  (Chapter  2, 
Section  9). 

J  Insulating  material  assumed  to  be  air,  the  permeability  of  which  is  unity. 


APPENDIX   B  313 

and  the  flux  per  centimeter  length  of  A  due  to  the  current  in  A  is  found  by  in- 
tegrating equation  (15)  between  the  limits  x  =  D  —  r  and  x  =  r. 

D-  r 

dx  ,  ,. 

-  (16) 

=  2  i  loge  — —  c.g.s.  units.  (17) 

The  total  flux  linking  with  the  current  in  conductor  A  is,  therefore, 


=  -  +  2  i  loge  —  '-  -c.g.s.  units.  (19) 

2  r 


From  equation  (4), 


L  =  *  («) 

=  -  -f-  2  loge  —    —  c.g.s.  units  *  per  centimeter  length 
of  conductor.  (21) 

2.  Inductance  of  each  conductor  in  a  three-phase  system.  —  The  inductance 
of  a  conductor  in  a  three-phase  system  depends  on  its  position  relatively  to  the 
other  conductors  of  the  system.  There  will  be  considered  here  only  the  two 
commonly  used  arrangements:  (a)  when  the  conductors  are  placed  at  the  ver- 
tices of  an  equilateral  triangle,  (b)  when  the  conductors  are  in  the  same  plane. 

(a)  When  the  conductors  are  placed  at  the  vertices  of  an  equilateral  triangle.  — 
In  a  balanced  three-phase  system,  the  currents 
in  A,  B  and  C  (Fig.  275a)  are  120  degrees  out  Jj* 

of  phase,  the  algebraic    sum   of    the    currents  //|\\ 

is  zero  (Chapter  7,   Section  3),  and  the  instan-  P/    \  \  V 

taneous  flux  set  up   around  any  conductor   is  //  ^A^  \\ 

the  algebraic  sum  of  the  instantaneous  fluxes  set  *&~          ~  ~^c 

up  around  the  other  two  conductors.    Therefore,  •-—-&----.. 

the  inductance  of  each  conductor  in  a  three-phase 
system,  when  the  conductors  are  placed  at  the 

vertices  of  an  equilateral  triangle,  is  equal  to      ^ZZILrz^~_IZ~lLZ 
one  half  the  inductance  of  the  loop  formed  by 
any  two  of  the  conductors,  and  is  calculated  by  F 

means  of  equation  (21). 

(6)  When  the  conductors  are  in  the  same  plane.  —  From  equation  (21)  the 
inductance  of  either  A  or  C  (Fig.  275b)  with  regard  to  B  is 


*  To  reduce  c.g.s.  units  of  inductance  to  henries,  divide  by  io9.  The  expression 
log*  —  -^-  may  be  replaced  by  loge  —  without  appreciable  error  when  the  ratio  —  is 
large. 


314  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

the  inductance  of  A  with  regard  to  C  or  of  C  with  regard  to  A  is 

^,  (33) 


and  the  inductance  of  B  with  regard  to  either  A  or  C  is 

D  —  r 


f    N 

(24) 

It  is  common  practice  to  transpose  the  conductors  of  polyphase  alternating- 
current  systems,  thus  making  the  inductances  of  the  lines  sensibly  equal. 
Under  this  condition  the  inductance  of  each  conductor  in  a  three-phase 
system,  when  the  conductors  are  in  the  same  plane,  is 


1  ,     ,       i.26(D—  r) 

=  -  -f-  2  loge  —  —  c.g.s.  units  per  centimeter  length 

2  r 

of  conductor.*  (26) 

3.  Inductance  of  a  conductor  with  earth  return.  —  A  consideration  of  Fig.  272 
shows  that  the  flux  linking  with  the  current  in  each  of  two  parallel  wires  sepa- 
rated D  centimeters  from  each  other,  passes  between  the  axis  of  the  wire  and 

a  neutral  plane  distant  -  centimeters  from  the  axis  of  the  wire.    When  the 

return  current  of  a  circuit  flows  through  the  earth,  the  surface  of  the  earth  is 
the  neutral  plane,  and  the  inductance  of  a  conductor  placed  h  centimeters 
above,  and  parallel  to,  the  surface  of  the  earth  is 

L  =  -  +  2  loge  —      —  c.g.s.  units  per  centimeter  length  of 
2  r 

conductor.  (27) 

Note.  —  Equation  (27)  is  strictly  true  only  when  the  resistance  of  the  earth 
return  is  zero,  but  a  considerable  resistance  in  the  return  circuit  does  not  greatly 
increase  the  inductance. 

4.  Inductance  of  a  solenoid.  —  The  inductance  of  a  solenoid,  either  with  or 
without  an  iron  core,  is  easily  calculated  when  it  is  assumed  that  the  flux  set  up 
by  the  coil  is  confined  to  the  interior  of  the  coil.|    Let 

N  =  the  number  of  turns  in  the  coil, 
i  =  the  current  flowing  in  the  coil,  in  c.g.s.  units, 
A  =  the  internal  cross-sectional  area  of  the  coil,  in  square  centimeters, 
/  =  the  length  of  the  coil,  in  centimeters, 
n  =  the  permeability  of  the  magnetic  circuit. 

The  total  flux  set  up  by  the  coil  (Chapter  2,  Section  10)  is 

(2g) 

*  Equation  (26)  gives  values  only  slightly  greater  than  those  given  by  equation  (21). 
t  This  condition  is  only  approximated  in  commercial  apparatus. 


APPENDIX  B 
and  the  inductance  of  the  helix  is 


c..s.  units. 


315 

(29) 
(30) 


b  +  c  + 


milhenries, 


(31) 


Since  the  permeability  of  iron  decreases  as  the  flux  density  increases,  the 
inductance  of  a  coil  wound  on  an  iron  core  is  not 
constant,   but  decreases   as   the   current   in  the  coil 
increases. 

5.  Inductance  of  a  coil.  —  Professors  Brooks  and 
Turner  *  have  developed  an  empirical  formula  by  means 
of  which  the  inductance  of  any  closely  wound  cylin- 
drical coil  (Fig.  276)  may  be  calculated.  The  error 
involved  in  the  use  of  this  formula  seldom  exceeds  4  per 
cent  for  a  coil  of  any  dimensions,  and  becomes  less  as 
the  relative  length  of  the  coil  increases. 


FIG.  276. 


when  a  =  the  mean  radius  of  the  winding,  in  centimeters, 
b  =  the  axial  length  of  the  coil,  in  centimeters, 
c  =  the  thickness  of  the  winding,  in  centimeters, 
R  —  the  outer  radius  of  the  winding,  in  centimeters, 
N  =  the  total  number  of  turns  in  the  winding, 


F' 

F 


106+  i2c-\-  2R 
lob 


ioc-f- 1. 
"  =  o.5logi0(  100  + 


14  R 


6.   Energy  stored  in  a  magnetic  field.  —  In  any  circuit  of  inductance  L, 

rdi 


PL 


e 

T.di 

=  Li  — 

dt 


(32) 
(33) 
(34) 


and  the  energy  required  to  increase  or  decrease  the  intensity  of  the  magnetic 
field  is 

WL  =  fpdt  (35) 

=  Lfidi  (36) 

(37) 


-X*. 

2 


when  the  current  varies  between  i  and  zero. 

*  University  of  Illinois  Bulletin,  Vol.  IX,  No.  10. 


316  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

Substituting  the  value  of  L  found  in  equation  (30) 


But  the  intensity  of  the  magnetic  field  (Chapter  2,  Section  9)  is 


I 

and  the  volume  of  the  magnetic  field  is 

V  =  Al  cubic  centimeters. 
Substituting  the  values  of  H  and  V  in  equation  (38) 


WL  =         X  V  ergs 

STT 

BH 

=  —  X  V  ergs 

8  7T 

Bff 

=  —  —  ergs  per  cubic  centimeter 

8  7T 


B* 

8  7TU 


ergs  per  cubic  centimeter. 


(38) 

(39) 
(40) 

(41) 
(42) 
(43) 
(44) 


7.  Loss  of  energy  due  to  hysteresis. — When  the  magnetic  circuit  is  composed 
wholly  or  partly  of  iron,  it  has  been  shown  that :  (a)  the  energy  delivered  to  the 

magnetic  field  is  not  all  returned  to  the 
system  when  the  magnetizing  force  is 
withdrawn,  (b)  that  the  flux  density  pro- 
duced by  a  given  magnetizing  force  is 
greater  for  decreasing  values  of  field 
intensity  than  for  increasing  values,  (c) 
that  the  iron  is  heated  when  the  flux 
density  changes. 

FIG.  277.    Hysteresis  Loop  and  Mag-         When  the  flux  density  B  in  the  iron 
netizing  Current  Wave.  and  the  magnetizing  force  H  of  a  mag- 

netic circuit  are  plotted  for  a  complete  magnetic  cycle,  i.e.,  for  all  possible 
values  between  a  given  positive  maximum  and  an  equal  negative  maximum 
density,  a  curve  similar  to  Fig.  277  is  obtained.  Let 

N  =  the  number  of  turns  in  the  exciting  coil  of  an  electromagnet, 

i  =  the  current  in  c.g.s.  units  flowing  in  the  coil, 

V  =  the  volume  of  the  iron  core  (/  centimeters  in  length  and  A  square 
centimeters  in  cross-sectional  area), 

d>  =  the  flux  in  the  iron  core. 


Then 


ei(df)  =  Ni 

eit  =  Ni 


(45) 

(46) 
(47) 


But 


and 


Therefore, 


APPENDIX  B 


<t>  =  AB 

HI 

Ni  = 

47T 

Wh  =  —  ( 

A-JT  J  - 


+B 


HdBc.g.s.  units, 


317 

(48) 

(49) 

(50) 
(51) 


i.e.,  the  energy  expended  in  carrying  a  volume  of  iron  through  a  magnetic 
cycle  is  proportional  to  the  area  of  the  hysteresis  loop,  and  the  average  power 
loss  due  to  hysteresis  in  the  iron  is 

i_  n 

(52) 


when  /  =  the  number  of  magnetic  cycles  per  second  through  which  the  iron 
passes. 

Dr.  C.  P.  Steinmetz  has  shown  that  the  area  of  a  hysteresis  loop  is  approxi- 
mately proportional  to  the  1.6  power  of  the  maximum  flux  density  attained 
during  a  magnetic  cycle.  The  loss  per  cycle  per  cubic  centimeter  of  iron  due 
to  magnetic  hysteresis  is,  then, 

(53) 


when  ij  =  the  hysteretic  (magnetic)  constant  and  is  dependent  on  the  physical 
properties  of  the  iron.    Table  XX. 

TABLE  XX 

HYSTERETIC  CONSTANTS  (r) 


Kind  of  iron 

Constant 

Best  annealed  sheet  

o  0015 

Good  annealed  sheet          

o  003* 

Ordinary  annealed  sheet  

o  004 

Soft  annealed  cast  iron  

o  008 

Soft  machine  steel  

O.OI 

Cast  steel  

0.  12 

Cast  iron  

o.  16 

Hardened  steel  

0.25 

*  Largely  used  for  dynamo  armature  punchings. 

8.  Distortion  of  current  wave  due  to  hysteresis.  —  From  Section  7  it  is  evi- 
dent that  the  magnetizing  current  in  a  coil  having  an  iron  core  in  which  the 
flux  varies  harmonically  cannot  be  harmonic,  but  has  a  distorted  wave  shape 
as  indicated  in  Fig.  277. 

9.  Growth  and  decay  of  current  in  an  inductive  circuit.  —  When  an  electro- 
motive force  of  constant  value  is  applied  to  a  circuit  containing  both  resist- 
ance and  inductance,  the  current  does  not  immediately  rise  to  its  final  value 

E« 

— ,  because  of  the  counter-electromotive  force  induced  in  the  circuit  by  the 


318  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

increasing  flux;  when  a  circuit  containing  both  resistance  and  inductance  is 
disconnected  from  a  source  of  constant  electromotive  force,  and  the  circuit 
closed,  the  current  does  not  immediately  fall  to  zero,  but  is  maintained  by  the 
electromotive  force  induced  in  the  circuit  by  the  decreasing  flux.  Let 

E  =  the  applied  electromotive  force, 
R  =  the  resistance  of  the  circuit, 
L  =  the  inductance  of  the  circuit, 

—  =  the  rate  at  which  the  current  flowing  in  the  circuit  changes. 

For  an  increasing  current, 

*•  (54) 


Transposing  equation  (54)  and  multiplying  by  —  R, 

-  R  (di)      -  R  (dl) 


E-Ri  L      ' 


and 


_ 

Ki  Li 


R  ^      -       »  (58) 


<-|(, -,-*)• 

For  a  decreasing  current, 

1  =  o,  (59) 


(6o) 


and 

i=|r^.  (63) 


APPENDIX  C 
CAPACITANCE 

1.  The  dielectric  field.  —  When  the  potential  of  a  body  is  greater  or  less  than 
zero  (the  surface  of  the  earth  is  assumed  to  be  at  zero  potential),  the  body  is  said 
to  be  electrified  or  charged.    The  energy  required  to  produce  electrification  is 
stored  in  the  surrounding  medium,  and  there  is  set  up  in  the  medium  a 
stressed  condition  termed  a  dielectric  or  electrostatic  field.    The  dielectric 
flux  emanating  from  a  surface  is  normal  to  the  surface  as  in- 
dicated in  Fig.  278,  and  is  represented  by  lines  which  termi- 
nate at  another  surface. 

2.  Properties  of  electric  charges. — The  following  properties 
of  electric  charges  have  been  established  by  experiment: 

(a)  Like  charges  repel;  unlike  charges  attract. 

(6)  When  a  charge  is  produced  on  any  body,  an  equal  and 
opposite  charge  is  produced  on  the  same  or  on  some  other 
body.  IG*  2^  * 

(c)  The  force  exerted  between  two  charges  of  electricity  is  directly  propor- 
tional to  the  product  of  the  charges,  and  inversely  proportional  to  the  square  of 
the  distance  separating  them. 

5      '         -        ,  /=^'  (I) 

3.  Electrostatic  units.  —  The  units  of  the  electrostatic  system,  and  the 
equivalent  values  in  practical  units  are:  . 

(a)  The  unit  of  quantity  (charge)  is  that  quantity  of  electricity  which  repels 
with  a  force  of  one  dyne  a  similar  and  equal  quantity  of  electricity  placed  at  a 
distance  of  one  centimeter  in  air.    To  reduce  electrostatic  units  of  quantity  to 
coulombs  divide  by  3  X  io9. 

(b)  Unit  current  is  that  which  conveys  unit  quantity  past  a  given  point  on 
a  conductor  in  one  second.    To  reduce  electrostatic  units  of  current  to  amperes 
divide  by  3  X  io9. 

(c)  Unit  electrostatic  force  or  unit  difference  of  potential  exists  between  two 
points  when  one  erg  of  work  is  expended  in  transferring  unit  quantity  of  elec- 
tricity from  one  point  to  the  other  against  the  force  of  the  electrostatic  field. 
To  reduce  electrostatic  units  of  potential  to  volts  multiply  by  300. 

(d)  Unit  capacitance  is  that  which  produces  unit  difference  of  potential  when 
charged  with  unit  quantity  of  electricity.    To  reduce  electrostatic  units  of  ca- 
pacitance to  microfarads  divide  by  900,000. 

(e)  The  specific  inductive  capacity  or  dielectric  constant  of  a  substance  is  the 
ratio  of  the  capacitance  of  a  condenser  having  that  substance  as  a  dielectric,  to 

319 


320 


ESSENTIALS  OF  ELECTRICAL  ENGINEERING 


the  capacitance  of  the  same  condenser  using  dry  air,  at  o°  C.  and  a  pressure  of 
76  centimeters,  as  the  dielectric.     (Table  XXI.) 

TABLE  XXI 
DIELECTRIC  CONSTANTS  (K) 


Material 

Constant 

Air  

I  .O 

Oil  (transformer)  

2.  I 

Shellac  

2.75 

Paraffin 

2    3 

Rubber 

2    1< 

Paper 

2  to    4 

Gutta  percha 

3  to    "? 

Glass                                               .    ... 

3  to  10 

Mica                          .       

4  to    8 

Varnished  cambric  

4  to    6 

Pure  water        

80 

(/)  The  intensity  of  a  dielectric  field  is  the  ratio  of  the  total  flux,  when 
propagated  in  air,  to  the  area  of  the  surface  from  which  it  emanates,  and  is 
equal  to  the  force  in  dynes  at  the  point.*  (Symbol  F.) 

(g)  The  density  of  a  dielectric  field  is  the  product  of  the  field  intensity  and 
the  dielectric  constant.*  (Symbol  D.) 

4.  Flux  due  to  unit  charge.  —  The  force  with  which  a  unit  charge  acts  on 
another  unit  charge  distant  r  centimeters  in  air  is  (from  2  c  and  3  a) 

f=  ^  dynes  (2) 


(3) 


and  the  intensity  of  the  dielectric  field  is 

F  =  -2  lines  per  square  centimeter. 

But  the  surface  of  a  sphere,  the  diameter  of  which  is  2  r  centimeters  is 
square  centimeters.    Therefore,  the  total  dielectric  flux  emanating  from  unit 
charge  is 

^=1X4^2  (4) 

=  47rlines.  (5) 

5.  Potential  difference  between  points.  —  The  potential  difference  between 
points  is,  by  definition,  equal  to  the  work  in  ergs  done  when  unit  quantity 
of  electricity  is  transferred  from  one  point  to  the  other  against  the  force  of  the 
electrostatic  field,  and  is,  therefore,  the  line  integral  of  the  field  intensity  be- 
tween the  points. 

(6) 


*  Compare  with  the  corresponding  terms  used  for  the  magnetic  circuit  (Chapter  2). 


APPENDIX  C 


32I 


6.  Capacitance.  —  Capacitance  has  been  denned  (Chapter  i,  Section  12)  as 
the  ratio  of  the  quantity  of  electricity  displaced  (the  charge),  to  the  electro- 
motive force  producing  the  displacement. 

0 


C= 


(7) 


7.  Capacitance  of  parallel  plates.  —  Determine  the  capacitance  of  two  par- 
allel plates  (Fig.  278)  A  square  centimeters  in  area,  and  separated  by  /  centi- 
meters of  dielectric,  the  constant  of  which  is  K.  Let  Q  be  the  charge  on  each 
square  centimeter  of  the  positive  plate  and—  Q  the  charge  on  each  square  centi- 
meter of  the  negative  plate.  Then 

(8) 

(9) 
(10) 


F  =          lines  per  square  centimeter, 
K 


y 
K 


electrostatic  units, 


and 


(ix) 


TT 

=  —  -  electrostatic  units  per  square  centimeter       (12) 

47T/ 


A  IF 

=  —  electrostatic  units 

= j-  900,000  microfarads, 


(13) 
(14) 


i.e.,  the  capacitance  of  a  plate  condenser  is  directly  proportional  to  the  product 
of  the  area  of  plates  and  the  dielectric  constant,  and  inversely  proportional  to 
the  distance  between  the  plates. 

Commercial  condensers  are  made  up  of  a  large  number  of  sheets  of  tinfoil,  to 


FIG.  279.    The  Plate  Condenser. 


FIG.  280. 


obtain  the  required  area  of  the  plates,  connected  as  indicated  in  Fig.  279,  and 
separated  by  sheets  of  paraffined  paper. 

8.  Capacitance  of  concentric  cylinders.  —  Determine  the  capacitance  of  two 
concentric  cylinders  (Fig.  280),  the  radii  xi  and  x2  of  the  cylinders,  their  length 
/,  and  the  dielectric  constant  K  being  given.  Let  the  charge  per  centimeter 
length  of  the  positive  cylinder  be  Q  and  the  charge  per  centimeter  length  of  the 


322  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

negative  cylinder  be  —  Q.    Then  the  field  intensity  at  any  point  distant  r  centi- 
meters from  the  axis  of  the  cylinders  is 


=  -£  lines  per  square  centimeter.  (16) 

The  potential  difference  between  the  two  cylinders  is  found  by  integrating 
equation  (15)  between  the  limits  r  —  x\  and  r  =  x% 

E  =  ^Jn~r  (I7) 

^log,'5  (18) 

.A.  Xv 


and  C  =  — U (19) 


*r 

electrostatic  units  per  centimeter 


length  of  cylinders.  (20) 

9.   Capacitance  of  parallel  wires.  —  Determine  the  capacitance  of  two  parallel 
wires  (Fig.  281),  their  radii  r,  the  distance  between  their  centers  D,  their 
length  /  and  the  dielectric  constant  K  being  given.    Let  the  charge  per  centi- 
meter length  of  the  positive  wire  be  Q  and  the  charge  per  centimeter  length  of 

the  negative  wire  be  —  Q.    Then  the  field  in- 
x  ..............  p     "p-;™;;::;-;~:£|       tensity  at  any  point  distant  x  centimeters  from 

K  ___  I  ____  /j>\    the  axis  of  conductor  A  is 


and  the  potential  difference  between  A  and  B  is  found  by  integrating  equation 
(21)  between  the  limits  x  =  D  —  r  and  x  =  r. 


*  The  expression  loge  -      -  may  be  replaced  by  loge  —  without  appreciable  error 
when  the  ratio  —  is  large. 


APPENDIX  C  323 


K              r 
=  — electrostatic   units   per    centimeter 

4loge— ^ 

length  of  circuit.  (25) 

10.   Capacitance  of  each  conductor  in  a  three-phase  circuit. — The  capacitance 
of  a  conductor  in  a  three-phase  system  depends  on  its 
position  relatively  to  the  other  conductors  in  the  system.  ^JL* 

There  will  be  considered  here  only  the  two  commonly 
used  arrangements:  (a)  when  the  conductors  are  placed  p''/       \'D. 

at  the  vertices  of  an  equilateral  triangle,  (b)  when  the 
conductors  are  in  the  same  plane. 


(a)  When  the  conductors  are  placed  at  the  vertices  of  an 
equilateral  triangle. — From   equation   (22)   the  poten-  pIG>  282a. 

tial  difference  between  any  conductor  (Fig.  28 2a)  and 
the  neutral  plane  halfway  between  the  conductor  and  either  of  the  other 
conductors  is 

' 


and  the  capacitance  of  each  conductor  is 

c=^L 


electrostatic  units  per  centimeter 


length  of  conductor,  (28) 

i.e.,  the  capacitance  of  each  conductor  in  a  three-phase  system,  when  the  con- 
ductors are  placed  at  the  vertices  of  an  equi- 

K~~ 0. >*<•- D •*! 

QA  -~-'-£JB —        — @-c  lateral  triangle,  is  twice  the  capacitance  of  the 

P  ,  loop  formed  by  any  two  of  the  conductors. 

(b)  When  the  conductors  are  in  the  same  plane. 

From  equation  (22)  the  potential  difference  between  A  or  B  (Fig.  282b)  and 
their  neutral  plane  is 

,-,20,7)  —  r  /N 

£l  =  -^loge— —   .  (29) 

the  potential  difference  between  A  or  C  and  their  neutral  plane  is 

fi-'log.,  (30) 


324  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

and  the  potential  difference  between  B  or  C  and  their  neutral  plane  is 

£3  =  flog«^.  .  (30 

Under  the  common  practice  of  transposing  the  conductors  of  polyphase 
systems,  the  capacitances  of  the  conductors  are  sensibly  equal,  and  may  be 
calculated  by  means  of  the  following  equation: 

c  = 


D-r  ,   2O,       2D-r  . 

—  +flog'-7- 

(33) 


D-r  .      .       2D-r 

— h  2  loge — 

K 


electrostatic  units  per  centimeter 


length  of  conductor.*  (34) 

n.  Charging  current  in  a  three-phase  system.  —  From  equation  (16),  Chapter 
i,  the  current  due  to  the  capacitance  of  a  single-phase  line  is 

IC=2TT/CE.  (35) 

when  the  capacitance  is  concentrated.     But  the  voltage  between  the  terminals 

of  a  condenser  star-connected  to  a  three-phase  system  is  ~  times  the  line-to- 

V3 

line  voltage.    Therefore,   the  charging  current  due  to  the  capacitance  of  a 
three-phase  line  is 

r        4  iffCE  /•  ,\ 

Ic  =     J  .-    ,  (36) 

^3 

and  is    -^=  (=  1.15)  times  the  charging  current  flowing  in  the  circuit  formed 

^3 

by  any  two  of  the  conductors  when  the  voltage  between  lines  is  equal  to  the 
line-to-line  voltage  of  the  three-phase  system. 

12.   Energy  stored  in  the  dielectric  field.  —  When  the  intensity  of  a  dielectric 
field  increases  or  decreases,  a  current  flows  to  or  from  the  charged  body. 


and  q  =  Ce.  '  '      (38) 

Therefore,  *'  =  CT/  <39> 

at 

*  Equation  (33)  gives  values  only  slightly  gr  sater  than  those  given  by  equation  (27)  . 


APPENDIX  C  325 

But  Wc  =       pdt  (40) 

(41) 

(42) 

*    . 
ergs  (43) 


f  eidt 
Cfede 


when  the  applied  voltage  varies  from  e  to  zero. 

:    Substituting  in  equation  (43)  the  value  of  e  obtained  in  equation  (9),  and  that 

of  C  obtained  in  equation  (13) 

(44) 
(45) 

IT 

But  F  =  |  (46) 

and  Al=V  cubic  centimeters.     ,  (47) 

Therefore,  We  =  ^  X  V  ergs  (48) 


7}  17 

=  -  —  ergs  per  cubic  centimeter  (49) 

8  T 

7-J2 

=  -  —  ergs  per  cubic  centimeter.          (50) 

8  7TA 

13.  Dielectric  hysteresis.  —  It  has  been  experimentally  shown  that   the 
energy  (heat)  dissipated  in  the  dielectric  of  a  condenser  is  greater  with  alter- 
nating current  than  when  a  constant  difference  of  potential  exists  between  the 
condenser  terminals,  the  effective  value  of  the  alternating  electromotive  force 
being  equal  to  the  constant  electromotive  force.    This  additional  loss  is  due 
to  so-called  dielectric  hysteresis. 

When  a  condenser  is  connected  to  an  alternating-current  circuit,  the  chang- 
ing value  of  the  dielectric  flux  induces  alternating  electromotive  forces  in  any 
conducting  particles  that  may  have  become  imbedded  in  the  dielectric.  Di- 
electric hysteresis  is,  therefore,  more  nearly  analogous  to  eddy  currents  than  to 
magnetic  hysteresis.  Dielectric  hysteresis  losses  are  always  small,  and  are 
usually  neglected. 

14.  Charging  and  discharging  current  in  a  condenser  circuit.  —  When  an 
electromotive  force  of  constant  value  is  applied  to  a  circuit  containing  both 
resistance  and  capacitance,  a  charging  current  flows  in  the  circuit;  when  a 
circuit  containing  both  resistance  and  capacitance  is  disconnected  from  a 


326  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

source  of  constant  electromotive  force,  and  the  circuit  closed,  a  discharging 
current  flows  in  the  circuit.     Let 

E  =  the  applied  electromotive  force, 
R  =  the  resistance  of  the  circuit, 
C  =  the  capacitance  of  the  circuit, 

-^  =  the  rate  at  which  the  capacitance  charges  or  discharges. 

For  a  charging  current, 

j~ 

(Si) 

(52) 


qq-CE-~JtRC'  (S3) 

-CE\          _^ 
-  CE  I          RC  (54) 


(55) 
But  CE  =  Q  (56) 

and  *  =  J'-  (57) 

n  I  -- M 

Therefore  *  =  lL\e   RC). 

RCV 

For  a  discharging  current, 


q  =  Qe'RC  ^  (63) 

and  n  ~RC 

*=-RC<      • 

15.  Corona.  —  The  phenomena  known  as  corona  *  is  an  electrostatic  (leak- 
age) discharge  between  wires  of  different  potential,  and  takes  place  when  the 

*  See  The  Transactions  of  the  American  Institute  of  Electrical  Engineers,  Vol. 
XXX,  pages  1889-1965  and  Vol.  XXXI,  pages  1051-1092,  "Law  of  Corona  and  the  Di- 
electric Strength  of  Air,"  by  F.  W.  Peek,  Jr.,  Vol.  XXXI,  pages  1035-1049,  "Corona 
Losses  Between  Wires  at  High  Voltages,"  by  C.  Francis  Harding. 


APPENDIX  C  327 

potential  difference  between  the  wires  exceeds  a  certain  "critical"  value,  de- 
pending on  the  diameters  of  the  wires  and  their  distance  apart.  In  an  alter- 
nating-current system,  the  power  losses  due  to  corona  are  proportional  to  the 
frequency  and  to  the  square  of  the  increase  in  voltage  above  the  critical  value, 
but  are  usually  negligible  for  voltages  up  to  45,<xx>.  At  voltages  materially 
higher  than  the  critical  value,  a  visible  halo-like  envelope  surrounds  the  con- 
ductor, the  diameter  of  the  envelope  increasing  as  the  voltage  increases. 


APPENDIX  D 
THE   COMPLEX   QUANTITY 

ADMITTANCE,  CONDUCTANCE  AND  SUSCEPTANCE 

1.  The  complex  quantity.  —  It  is  evident  that  the  equation 

a  =  Vb*  +  ?  (i) 

may  be  written 

a  =  b  ±jc,  (2) 

when  it  is  specified  that  j  indicates  that  b  and  c  are  at  right  angles,  and  are  to 
be  combined  geometrically. 

j  may  be  defined  as  the  complex  operator,  the  effect  of  which  is  to  rotate  a 
vector  to  which  it  is  applied,  90  degrees  forward  in  relation  to  the  reference 
line.  The  effect  of  j  Xj  or  j*  is  to  rotate  the  vector  180  degrees  in  relation 
to  the  reference  line,  and  the  algebraic  value  of 

f  —  i.  (3) 

Therefore,  j=  V—  i.  (4) 

Addition,  subtraction,  multiplication  and  division  of  equations  involving  the 
complex  quantity  are  essentially  algebraic  operations,  and  are  governed  by 
the  laws  of  such  processes.  The  use  of  the  complex  quantity  greatly  simplifies 
the  solution  of  some  alternating-current  problems,  particularly  those  relating 
to  transmission  lines  and  distributing  networks. 

2.  Examples.  —  (i)  Find  the  line  current  when  the  currents  in  the  parallel 
branches  of  a  circuit  are:  /i  =  6  —  j  3,  and  /2  =  8  —  j  2. 


=  14.86  amperes. 

(2)  Find  the  applied  electromotive  force,  the  power  component,  the  wattless 
component,  and  the  power  in  a  series  circuit  when:  /  =  5  +  ^*2  and 

£=20+.;  5. 
E  =  ZI 

=  (20+75)  (5  +72) 

=  100  +.725+.;  40  +/  10 

=  90+^65  (since/  =  -  i) 

=  V(9o)*+(65)2 
=  in  volts  (applied). 
EI  =  90  volts  (power  component). 
328 


APPENDIX  D  329 

1 

E2  =  65  volts  (wattless  component). 
El  =  (90  +./  65)  (5  -K;  2) 


=  320+^505 
P  =  320  watts. 

(3)  Find  the  power  in  an  electric  circuit  when:  I  =  10  -f  j  4  and 

E  =  150  —j  24. 
El  =  (150  -  j  24)  (10  +j  4) 
/  1          =  1  500  —  j  240  +  j'  600  —  j2  96  - 

=  1596  +.7360, 
P  =  1596  watts. 

3.  Admittance,  conductance  and  susceptance.  —  The  value  of  the  current  in 
any  circuit  is  expressed  by  the  equation 


'-f 


R±jX 
Multiplying  the  right-hand  member  of  equation  (7)  by  R  =F  jX 


(6) 
(7) 


The  value  of  the  current  in  the  circuit  may  also  be  expressed  by  the  equation 

7  =  YE  (10) 

=  (g*jQE,  (11) 

when    F  =  —  and  is  termed  the  admittance  of  the  circuit, 

£ 

g  —  Y  cos  </>  and  is  termed  the  conductance  of  the  circuit, 
b  =  Y  sin  <f>  and  is  termed  the  susceptance  of  the  circuit. 

From  equations  (n)  and  (9) 


i.e.,  the  conductance  of  a  series  circuit  is  equal  to  its  resistance  divided  by  the 
square  of  its  impedance 


*  = 


Compare  equations  (13)  and  (14)  with  equations  (124)  and  (129),  Chapter  I. 


330  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

and  the  susceptance  of  a  series  circuit  is  equal  to  its  reactance  divided  by  the 
square  of  its  impedance. 


Admittances,  like  impedances,  must  be  combined  geometrically;   conduc- 
tances or  susceptances  are  combined  algebraically. 

APPENDIX  D  —  PROBLEMS 

1-6.   Solve  Problems  i  to  6,  Chapter  18,  using  Complex  Quantity. 
7-14.  Solve  Problems  22,  30,  34,  35,  36,  40,  41  and  42,  Chapter  i,  using  ad- 
mittance, conductance  and  susceptance  values. 

*  Compare  equations  (13)  and  (14)  with  equations  (120)  and  (125),  Chapter  i. 


APPENDIX  E 

RULES  RECOMMENDED  BY  COMMISSION  ON  RESUSCITATION 
FROM  ELECTRIC   SHOCK 

REPRESENTING 

The  American  Medical  Association 

The  National  Electric  Light  Association 

The  American  Institute  of  Electrical  Engineers 

DR.  W.  B.   CANNON,  Chairman  DR.   GEORGE  W.   CRILE 

Professor   of  Physiology,    Harvard  Professor  of  Surgery,  Western  Reserve 

University  University 

DR.  YANDELL  HENDERSON  MR.  W.   C.  L.  EGLIN 

Pr of  essor  of  Physiology,  Yale  Univer-  Past-President,     National     Electric 

sity  Light  Association 

DR.  S.  J.  MELTZER  DR.  A.  E.   KENNELLY 

Head  of  Department  of  Physiology  Professor  of  Electrical  Engineering, 

and  Pharmacology,  Rockefeller  In-  Harvard  University 

stitute  for  Medical  Research  DR.   ELIHU  THOMSON 

DR.  EDW.  ANTHONY  SPITZKA  Electrician,    General   Electric    Com- 

Director  and  Professor  of  General  pany 

Anatomy,  Daniel  Baugh  Institute  of  MR.    W.    D.    WEAVER,    Secretary 

Anatomy,  Jejferson  Medical  College  Editor,  Electrical  World 

Copyright,  1912,  by 
NATIONAL  ELECTRIC  LIGHT  ASSOCIATION 

Follow  these  instructions  even  if  victim  appears  dead. 

I.   IMMEDIATELY  BREAK  THE   CIRCUIT 

With  a  single  quick  motion,  free  the  victim  from  the  current.  Use  any  dry 
non-conductor  (clothing)  rope,  board,  to  move  either  the  victim  or  the  wire. 
Beware  of  using  metal  or  any  moist  material.  While  freeing  the  victim  from 
the  live  conductor  have  every  effort  also  made  to  shut  off  the  current  quickly. 

H.  INSTANTLY  ATTEND  TO  THE  VICTIM'S  BREATHING 

i.  As  soon  as  the  victim  is  clear  of  the  conductor,  rapidly  feel  with  your  finger 
in  his  mouth  and  throat  and  remove  any  foreign  body  (tobacco,  false  teeth, 
etc.).  Then  begin  artificial  respiration  at  once.  Do  not  stop  to  loosen  the  vic- 
tim's clothing  now;  every  moment  of  delay  is  serious.  Proceed  as  follows: 


332  ESSENTIALS  OF  ELECTRICAL  ENGINEERING 

(a)  Lay  the  subject  on  his  belly,  with  his  arms  extended  as  straight  forward 
as  possible  and  with  face  to  one  side,  so  that  nose  and  mouth  are  free  for  breath- 
ing (see  Fig.  i).  Let  an  assistant  draw  forward  the  subject's  tongue. 


FIG.  1.  —  Inspiration;  pressure  off. 

(6)  Kneel,  straddling  the  subject's  thighs,  and  facing  his  head;  rest  the  palms 
of  your  hands  on  the  loins  (on  the  muscles  of  the  small  of  the  back),  with  fingers 
spread  over  the  lowest  ribs,  as  in  Fig.  i. 

(c)  With  arms  held  straight,  swing  forward  slowly  so  that  the  weight  of  your 
body  is  gradually,  but  not  violently,  brought  to  bear  upon  the  subject  (see  Fig. 
2).     This  act  should  take  from  two  to  three  seconds. 

(d)  Then  immediately  swing  backward  so  as  to  remove  the  pressure,  thus 
returning  to  the  position  shown  in  Fig.  i. 

(e)  Repeat  deliberately  twelve  to  fifteen  times  a  minute  the  swinging  for- 
ward and  back  —  a  complete  respiration  in  four  or  five  seconds. 

(/)  As  soon  as  this  artificial  respiration  has  been  started  and  while  it  is  being 
continued,  an  assistant  should  loosen  any  tight  clothing  about  the  subject's 
neck,  chest,  or  waist. 

2.  Continue  the  artificial  respiration  (if  necessary,  two  hours  or  longer), 
without  interruption,  until  natural  breathing  is  restored,  or  until  a  physician 
arrives.  If  natural  breathing  stops  after  being  restored,  use  artificial  respira- 
tion again. 


FIG.  2.  —  Expiration;  pressure  on. 

3.  Do  not  give  any  liquid  by  mouth  until  the  subject  is  fully  conscious. 
Give  the  subject  fresh  air,  but  keep  him  warm. 

III.   SEND  FOR  NEAREST  DOCTOR  AS  SOON  AS  ACCIDENT 
IS  DISCOVERED 


REFERENCES 


"Alternating  Currents,"  F,  Bedell  and  C,  A.  Crehore. 

"Alternating  Currents,"  A.  Russell. 

"Alternating  Currents  and  Alternating  Current  Machinery,"  D.  C,  and  J.  P  Jackson, 

"Alternating  Current  Motors,"  A.  S.  McAllister. 

"Alternating  Current  Transformer,"  J.  A.  Fleming. 

"Alternating  Current  Windings,"  C.  Kinzbrunner. 

"Continuous  Current  Armatures,"  C.  Kinzbrunner. 

"Dynamo-electric  Machinery,"  S,  P.  Thompson. 

"  Electrical  Conductors,"  F.  A,  C.  Perrine. 

"Electrical  Meters,"  C.  M.  Jansky. 

"  Electrical  Illuminating  Engineering,"  W.  E.  Barrows,  Jr. 

"Electric  Lighting,"  F.  B.  Crocker. 

"Electric  Motors,"  F.  B.  Crocker  and  M.  Arendt. 

"Electric  Power  Conductors,"  W.  A.  del  Mar. 

"Electric  and  Magnetic  Calculations,"  A.  A.  Atkinson. 

"Electric  Transmission  of  Energy,"  A.  V.  Abbott. 

"Elements  of  Electrical  Engineering,"  C.  P.  Steinmetz. 

"Elements  of  Electrical  Transmission,"  O.  J.  Ferguson. 

"  Experimental  Electrical  Engineering,"  V.  Karapetoff. 

"  Foster's  Electrical  Engineers  Handbook." 

"Illumination  and  Photometry,"  W.  E.  Wickenden. 

"  Polyphase  Apparatus,"  M.  A.  Oudin. 

"Polyphase  Alternating  Currents,"  S.  P.  Thompson. 

"Principles  of  Electrical  Engineering,"  H.  Pender. 

"  Secondary  Batteries,"  E.  J.  Wade. 

"Standard  Handbook  for  Electrical  Engineers." 

"'Synchronous  Motors  and  Converters,"  A.  E.  Blondel. 

"The  Electric  Circuit,"  V.  Karapetoff. 

"The  Induction  Motor,"  B.  F.  Bailey. 

"Transformer  Practice,"  W.  T.  Taylor. 


333 


INDEX 


Admittance,  329. 

Abvolt,  33. 

Air-break  switches,  244. 

All-day  efficiency,  73,  189. 

Alternating-current  circuits,  9,  10,  n,  13,  14,  270. 

resonance  in,  22. 

Alternating  current,  hydraulic  analogy  for,  2. 
Alternating-current  series  motor,  227. 
Alternator  efficiency,  143. 
Alternator,  frequency  of,  56. 
Alternator,  inductor,  57. 
Alternator  losses,  141. 
Alternator  ratings,  146. 
Alternator  regulation,  134. 
Alternator,  single-phase,  127. 

speed  and  frequency  of,  56. 

three-phase,  128. 

two-phase,  128. 

voltage  of,  123. 

Alternators  in  parallel,  load  division  of,  144,  158. 
Alternators,  parallel,  operation  of,  143. 
synchronizing  of,  144. 
voltage  characteristics  of,  133. 
American  wire  gauge,  269. 
Ammeters,  252. 
Ammeter  shunts,  254. 
Ampere,  4. 

Ampere-turn,  relation  of  the  gilbert  and  the,  39. 
Armature  core,  46. 

Armature  inductance,  compensation  for,  229. 
Armature  reaction,  64,  81,  129. 
Armature  windings,  basket,  51. 
chain,  51. 
distributed,  124. 
lap,  49. 
wave,  50. 
Arc  lamps,  carbon,  238. 

flaming,  238. 

magnetite,  237. 

Automatic  starting  rheostats,  90. 
Auto- transformer,  the,  190. 
Average  value  of  the  sine,  307. 
Average  value  of  the  sin2,  307. 

Balanced  three-phase  system,  109. 
Batteries,  299-304. 
Benchboards,  251. 
BK  motor,  Wagner,  232. 
Boosters,  303. 
Booster  converter,  170. 
Brown  and  Sharpe  wire  gauge,  269. 
Brush  advance,  64. 
Brushes,  carbon,  53. 
copper,  53. 

335 


336  INDEX 

Brush-contact  losses,  94. 
Brush  holders,  54. 
Building  up,  61,  70. 

Capacitance,  8,  320. 

of  concentric  cylinders,  321. 
'of  parallel  plates,  320. 
of  parallel  wires,  322. 
of  three-phrase  circuits,  323. 
of  transmission  lines,  270. 
Carbon-break  switches,  244. 
Carrying  capacity  of  wires,  270. 
C.  G.  S.  unit  of  current,  39. 

Circuits,  alternating-current,  9,  10,  n,  13,  14,  270. 
parallel,  3. 
series,  2. 
Circle  diagram  for  induction  motor,  214-217. 

for  synchronous  motor,  156. 
Circular  mil,  5. 
Charge,  electric,  319. 
Charging  current,  270,  324. 
Coefficient,  leakage,  41. 

temperature,  5. 

Commutation,  continuous-current  dynamo,  62-66,  81. 
series  alternating-current  motor,  227. 
sparkless,  64.     , 
Commutation,  resistance,  61. 

voltage,  61. 
Commutating  flux,  64. 
Commutator,  52,  161. 
Comparison  of  series  motors,  229. 
Comparison  of  star-  and  delta-connected  systems,  113. 
Comparison  of  two-  and  three-phase  systems,  119. 
Compensated  repulsion  motor,  231. 
Compensation  for  armature  inductance,  229. 
Compensation,  voltmeter,  282. 
Complex  quantity,  328. 
Compound  dynamo,  55. 
generator,  69. 
motor,  86,  87. 
Compounding,  70,  71. 
Concatenation  of  induction  motors,  225. 
Concentric  cylinders,  capacitance  of,  321. 
Conductance,  329. 
Conductors,  268. 
Connection  to  load  circuits,  76. 
Constant-current  transformer,  the,  192. 
Constancy  of  power  in  polyphase  systems,  115. 
Constants,  hysteretic,  317. 
Construction  of  transformers,  183. 
Continuous-current  dynamo,  54. 
Continuous-current,  hydraulic  analogy  for,  2. 
Control,  speed,  82. 
Cooling  of  transformers,  188. 
Copper  loss  in  transformers,  186. 
Core,  armature,  46. 
Corona,  326. 
Coulomb,  4. 

Counter- electromotive  force,  40,  79,  150,  178. 
Cross-magnetizing  magnetomotive  force,  65,  131. 
Cumulative  compound  motor,  86,  103. 
Current,  charging,  270,  324. 
C.  G.  s.  unit  of,  39. 


INDEX  337 


Current,  density  in  carbon  brushes,  61. 
effective,  17. 
induced,  32. 

relations  in  transformers,  180. 
the  electric,  i. 
triangle,  14. 
transformer,  255. 
wattless,  22,  23,  173.       , 

Decay  of  current  in  an  inductive  circuit,  318. 

Delta-connected  three-phase  system,  in. 

Demagnetizing  magnetomotive  force,  66. 

Detector,  ground,  266. 

Determination  of  stray  power,  experimental,  98. 

Diagram,  Mershon's,  278. 

Dielectric  field,  319. 

energy  stored  in,  324. 
flux,  319. 
hysteresis,  235. 

Differential  compound  motor,  87. 
Direction  of  magnetic  flux,  29,  30. 
Distributed  armature  windings,  124. 

effects  of,  126. 

Distribution  of  flux  in  air  gap,  64. 
Discharging  current  in  condenser  circuit,  326. 
Distributing  systems,  3,  273. 
Distortion  of  current  wave  due  to  hysteresis,  316. 
Dynamo,  compound,  55. 

elementary,  42. 

losses,  93. 

parts  of,  46-51. 

series,  55. 

separately  excited,  54. 

shunt,  54. 

Economy  of  the  three-phase  system,  119. 

Eddy  current  brake,  260. 

Eddy  currents,  95. 

Edison  battery,  301. 

Electric  arc,  instability  of  the,  240. 

power  factor  of  the,  240. 
Electric  charges,  properties  of,  319. 
Electric  current,  field  intensity  produced  by  the,  37. 

manifestations  of,  2. 
^Electric  currents,  i. 

hydraulic  analogy  for,  2. 
Electric  circuit,  power  in  an,  15. 
Electric  shock,  resuscitation  from,  331. 
Electric  units,  4. 
Electrical  degree,  125. 
Electrolytic  cell,  284. 
Electrodynamometer,  253. 
Electromagnet,  the,  30. 
Electrostatic  units,  319. 

voltmeter,  256. 
Effective  current,  17. 

electromotive  force,  17. 
Effects  of  distributed  armature  winding,  126. 
Efficiency,  101. 

all-day,  73. 

alternator,  143. 

synchronous  motor,  154. 

transformer,  188. 


338  INDEX 

Electrical  degree,  125. 

Electricity,  nature  of,  i. 

Electromotive  force,  4,  33,  60,  79,  123,  165,  180. 

Electromotive  force,  counter,  40,  79. 

effective,  17. 
Elementary  dynamo,  42. 
Elimination  of  harmonics,  126. 
Energy  stored  in  a  dielectric  field,  324. 
magnetic  field,  315. 
Equalizer,  50,  75. 
Equivalent  reactance,  187. 

resistance,  186. 

single-phase  system,  120. 
Experimental  determination  of  stray  power,  90. 

Faraday's  Law,  31. 
Feeder  regulation,  279. 
Field  discharge  resistance,  58. 
Field  intensity,  magnetic,  30. 
dielectric  320. 
due  to  unit  pole,  37. 
produced  by  an  electric  current,  37. 
Field,  magnetic,  29. 
poles,  46. 
windings,  46. 
Flux,  commutating,  64. 
density,  31. 
dielectric,  319. 
due  to  unit  charge,"  320. 

pole,  37. 
leakage,  41,  180. 
magnetic,  31. 
Force,  magnetizing,  31. 

magnetomotive,  31. 
of  magnetic  traction,  39. 
Frame,  46. 
Frequency,  57. 

changer,  221, 
Frequency  meters,  split  phase,  266. 

vibrating  reed,  265. 
Frequency  of  alternator,  56. 
Friction  and  iron  losses,  separation  of,  100. 
Frictional  losses,  97. 
Fundamental  equation  of  the  generator,  60. 

motor,  79. 

Fundamental  physical  action  in  the  transformer,  178. 
Fuses,  244. 

Gap,  spark,  257. 

General  Electric  RI  motor,  232. 

Generator,  compound,  69. 

fundamental  equation  of  the,  60. 

induction,  224. 

series,  68. 

shunt,  66. 

Generators,  parallel  operation  of,  73,  143. 
Gilbert  and  the  ampere-turn,  relation  of  the,  39. 
Graphic  meters,  262. 
Ground  detector,  266. 
Growth  of  current  in  an  inductive  circuit,  318. 

Harmonic  quantities,  305. 

combination  of,  305. 
examples  of,  305. 


INDEX  339 


Harmonic  quantities,  instantaneous  values  of,  305. 
rate  of  change  in,  305. 
resolution  of,  307. 
Harmonics,  elimination  of,  126. 
Heating,  104. 

High-voltage  measurements,  256. 
Holders,  brush,  54. 
Horn  gap,  284. 
Hot  wire  instruments,  252. 
Hunting,  154. 

Hydraulic  analogy  for  alternating  currents,  2. 
continuous  currents,  2. 
pulsating  currents,  2. 
Hysteresic  constants,  317. 
Hysteresis,  dielectric,  325. 

distortion  of  current  wave  due  to,  316. 
loop,  316. 
loss,  95,  316. 

Incandescent  lamps,  240. 
Inductance,  7. 

of  a  coil,  314,  315. 

of  a  conductor  with  earth  return,  314. 
of  a  three-phase  system,  313. 
of  parallel  wires,  311. 
of  transmission  lines,  270. 
Impedance,  12. 
Impedances  in  parallel,  20. 

series,  19. 

Induced  currents,  32. 
Induction  by  varying  flux  density,  35. 
instruments,  253. 
generator,  224. 

motor  action,  circle  diagram,  214. 
construction,  203. 
generator  action,  206. 
performance  curves,  213. 
slip  of  rotor,  206. 
slip-ring  rotor,  203. 
squirrel  cage  rotor,  203. 
starting,  210. 

synchronous  speed  of,  206-7. 
torque,  207. 
regulator,  280'. 
Impedance,  12. 

triangle,  12. 

Inductor  alternator,  57. 
Instability  of  the  electric  arc,  240. 
Instruments,  use  of,  23. 

electrodynamometer,  253. 
hot  wire,  252. 
induction,  253. 
permanent  magnet,  252. 
soft  iron,  253. 
Insulation,  wire,  269. 
Insulators,  pin,  269. 

suspension,  269. 
Interpoles,  64. 
Inverter  converter,  165. 
Iron  losses,  95. 

in  transformers,  184. 
separation  of,  TOO,  184. 
separation  of  friction  and,  100. 


340  INDEX 

Joule,  4. 
Joule's  Law,  15. 

Kilowatt,  5. 

Lamps,  arc,  235. 

incandescent,  240. 
nitrogen-filled,  241. 
Nernst,  241. 
mercury  vapor,  241. 
regenerative,  239. 
quartz,  242. 
Law,  Faraday's,  32. 
Joule's,  15. 
Lenz',  32. 

of  the  magnetic  circuit,  31. 
Ohm's,  6. 
Leakage  coefficient,  41. 

magnetic,  40,  180. 
Lightning  arresters,  283. 
Limitations  of  the  single-phase  system,  107. 
Limits,  regulation,  139. 
Line  calculations,  271. 

Load  circuits,  connection  of  generators  to,  76. 
Load  division  of  alternators,  144,  158. 
Losses,  brush-contact,  94. 

determination  of,  98,  141,  184,  186. 

dynamo,  93. 

eddy  current,  95. 

frictional,  97. 

in  alternator,  141, 

iron,  95. 

pole-face,  97. 

hysteresis,  95,  141,  183,  316. 

Magnetic  circuit,  law  of  the,  31. 
dampers,  158. 
field,  29. 

production  of  a,  29. 
energy  stored  in  a,  315, 
flux,  31. 

due  to  unit  pole,  37. 
leakage,  40,  180. 
neutral,  63,  64. 
traction,  force  of,  39. 
units,  30. 
Magnetism,  29. 
Magnetization  curves,  42. 
Magnetizing  force,  31. 
Magnetomotive  force,  30,  31. 
Manifestations  of  the  electric  current,  2. 
Maximum  load  of  synchronous  motor,  154. 
torque  of  induction  motor,  216, 
Measurements,  high-voltage,  256. 
Mercury  arc  rectifier,  construction,  161. 
efficiency,  163. 
limitations,  163. 
operation,  163. 
starting,  162. 
use,  164. 

Mercury  vapor  lamp,  241. 
Mershon's  diagram,  278. 
Meters,  graphic,  262. 


INDEX  341 


Meters,  power  factor,  264. 

recording,  262. 
Mil,  circular,  5. 
Moore  tube,  241. 
Motor,  alternating-current  series,  227. 

compensated  repulsion,  231. 

fundamental  equation  of,  79. 

generator,  161. 

General  Electric  RI,  232. 

operated  switches,  249. 

repulsion,  230. 

series,  85. 

shunt,  81. 

starting  rheostats,  87. 

Wagner  BK,  232. 
Multiple- wire  systems,  274. 
Multiplier,  256. 

Nernst  lamp,  241. 
Neutral  in  three-phase  system,  109. 
Neutral,  magnetic,  63,  64. 
Nitrogen-filled  lamp,  241. 

Oil-break  switches,  245. 
Ohm,  4. 
Ohm's  Law,  6. 
Oscillograph,  the,  126. 

Parallel  circuits,  3,  274. 
Parallel  operation  of  alternators,  143. 
generators,  73. 
induction  generators,  224. 
Parallel  wires,  capacitance  of,  322. 
inductance  of,  311. 
Parts  of  the  dynamo,  46-51. 
Permanent  magnet  instruments,  252. 
Permeability,  30,  31. 
Phase  characteristic,  155. 
difference,  307. 
transformation,  199. 
Pole-face  losses,  97. 
Poles,  field,  46. 

Polyphase  armature  reaction,  131. 
Polyphase  systems,  constancy  of  power  in,  115. 
power  factor  of,  114 
power  in,  114. 
Polyphase  transformers,  194. 

wattmeters,  260. 
Potential  difference  between  points,  319. 

transformer,  256. 

Properties  of  electric  charges,  319. 
Power  in  an  electric  circuit,  15. 

in  a  polyphase  system,  114. 
factor,  18. 
factor  meters,  264. 
factor  of  polyphase  system,  114. 
factor  of  the  electric  arc,  240. 
measurements,  116. 
Production  of  a  magnetic  field,  29. 
Proof  of  the  circle  diagram,  217. 
Pulsating  current,  hydraulic  analogy  for,  2. 
Protection  of  motors,  245. 


342  INDEX 

Protection  of  transformers,  196. 

Quarter-phase  system,  107. 
Quartz  lamp,  241. 

Ratings,  104,  146. 
Reactance,  10. 

equivalent,  187. 

synchronous,  136. 
Reactances  in  parallel,  19. 

series,  18. 
Reaction,  armature  64,  81,  129. 

of  magnetic  field  and  current-carrying  conductor,  36. 
Recording  meters,  262. 
Regulation,  alternator,  134. 

feeder,  279. 

limits,  139. 

speed,  82. 

transformer,  189. 

voltage,  61. 

Regulating-pole  converter,  169. 
Regulator,  induction,  280. 
Tirrill,  72,  140. 

Relation  of  the  gilbert  and  the  ampere-turn.,  39. 
Reluctance,  30,  31. 
Repulsion  motor,  230. 
Resistance,  5. 

and  reactance  in  parallel,  19. 

effect  of  temperature  on,  5. 

equivalent,  186. 

field  discharge,  58. 
Resistances  in  parallel,  19. 

series,  18. 
Resonance,  10. 

in  alternating-current  circuits,  22. 
Reversing  direction  of  induction  motor  rotation,  206. 

motor  armature  rotation,  36. 
Rheostats,  motor-starting,  87. 
RI  motor,  General  Electric,  232. 
Rotating  field  in  single-phase  induction  motor,  219. 
flux,  204. 

Saturation  curve  for  alternator,  134. 
Separately  excited  dynamo,  56. 
Separation  of  friction  and  iron  losses,  100. 
Series  circuits,  3,  273. 

dynamo,  55. 

generator,  68. 

motor,  85. 

on  alternating-current  circuit,  227. 
torque  of,  85. 

motors,  comparison  of,  229. 
Series-parallel  starting,  87. 
Series  transformer,  255. 
Short-circuit  curve  for  alternator,  135. 
Shunts,  ammeter,  254. 
Shunt  dynamo,  54. 

generator,  66. 

motor,  81. 

on  alternating-current  circuit,  226. 
Sine,  average  value  of,  307. 
Sin2,  average  value  of,  307. 
Single-phase  alternator,  127. 


INDEX  343 


Single-phase  armature  reaction,  129. 
induction  motor,  218. 

generator  action,  218. 
rotating  flux,  219. 
speed- torque  curves,  221. 
starting,  219. 
transformer  action,  218. 
system,  equivalent,  120. 

limitations  of,  107. 
Skin  effect,  271.  ' 

Slip  of  induction  motor,  206. 
Solenoid-operated  switches,  246. 
Solenoid,  the,  29. 
Soft  iron  instruments,  253. 
Spark  gap,  257. 
Sparking,  87,  104. 
Sparkless  commutation,  64. 
Speed  of  alternator,  56. 
Speed-torque  relations,  80. 
Speed  control,  82. 

regulation,  82. 
Split-pole  converter,  169. 
Stability  of  synchronous  motor,  153. 
Star-connected  three-phase  system,  109. 
Starting  of  synchronous  motors,  151. 
Storage  battery,  internal  actions  of,  299. 
Stray  power,  experimental  determination  of,  98. 
Stroboscopic  method  of  slip  measurement,  206. 
Structure  of  synchronous  motor,  149. 
Susceptance,  329. 
Switchboards,  249. 
Systems,  distributing,  273. 
Switches,  air-break,  244. 
automatic  245. 
carbon-break,  244. 
oil-break,  245. 
motor-operated,  249. 
solenoid-operated,  246. 
Switch  operation,  246. 
Synchroscope,  262. 

Synchronous  converter,  armature  reaction,  172. 
compounding,  169. 
construction,  164. 
current  relations,  167. 
efficiency,  171. 
heating,  173. 
hunting,  171. 
operation,  164. 
rating,  176. 
regulating-pole,  169. 
starting,  168. 
voltage  relations,  165. 
Synchronous  motor,  efficiency  of,  154. 

maximum  load  of,  154. 
stability  of,  153. 
starting  of,  151. 
structure  of,  149. 
torque-load  adjustment  of,  150. 
Synchronous  phase  modifier,  157. 
Synchronous  speed  of  induction  motor,  206-7. 
Synchronizing  of  alternators,  144. 

T-connected  three-phase  system,  112. 


344  INDEX 

Temperature  coefficient,  5. 
Terminal  voltage,  139. 
Three-phase  alternator,  128. 

system,  balanced,  109. 

capacitance  of,  323. 
delta-connected,  in. 
economy  of,  119. 
star-connected,  109. 
Three-phase  system,  T-connected,  112. 
unbalanced,  114. 
V-connected,  112. 
Tirrill  regulator,  72,  140. 
Torque  equation  for  induction  motors,  209. 
Torque-load  adjustment  of  synchronous  motors,  150. 
Torque  of  series  motor,  85. 
shunt  motor,  81. 
Traction,  force  of  magnetic,  39. 
Transformation,  phase,  199. 
Transformer,  fundamental  physical  action  in  a,  178. 

connections  for  single-phase  circuits,  196. 

synchronous  converters,  200. 
three-phase  circuits,  198. 
two-phase  circuits,  198. 
constant-current,  192. 
current,  255. 
losses,  184,  186. 
potential,  256. 
Transformers,  construction  of,  183. 

current  relations  in,  180. 
cooling  of,  188. 
efficiency  of,  188. 
polyphase,  194. 
protection  of,  196. 
series,  255. 
types  of,  183. 
Vector  diagrams  for,  180. 
Transmission  lines,  capacitance  of,  270. 
inductance  of,  270. 
Triangle,  current,  14. 

impedance,  12. 
voltage,  12. 
Tube,  Moore,  241. 
Two-phase  system,  107. 
Types  of  transformers,  183. 

Unbalanced  three-phase  system,  114. 
Units,  electric,  4. 

electrostatic,  319. 
magnetic,  30. 

Unit  of  current,  C.  G.  S.,  39. 
pole,  30. 

field  intensity  due  to,  37. 
magnetic  flux  due  to,  37. 
Use  of  instruments,  23. 

V-connected  three-phase  system,  112. 
Vector  diagrams  for  transformers,  180. 
Volt,  4,  33. 
Voltage  characteristics  of  alternators,  133. 

compound  generators,  70. 

series  generators,  68. 

shunt  generators,  67. 
Voltage  of  alternators,  123. 


INDEX  345 


Voltage  of  continuous-current  generators,  60. 

regulation,  61. 

terminal,  139. 

triangle,  12. 
Voltmeters,  252. 
Voltmeter  compensation,  282. 
electrostatic,  256. 

Wagner  BK  motor,  232. 

Watt,  4. 

Wattless  current,  22,  23,  173. 

Watt-hour  meter,  commutator,  260. 

induction,  261. 
Wattmeter  connections,  116. 
Wattmeters,  electrodynamometer,  258. 
induction,  258. 
polyphase,  260. 

Windings,  armature,  basket,  51. 
chain,  51. 
lap,  49. 
wave,  50. 
field,  46. 

Wires,  carrying  capacity  of,  270. 
Wire  gauge,  American,  270. 
Wiring  formulae,  276. 
Work,  15. 

Yoke,  52. 

Y-connected  three-phase  system,  109. 


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GOLDSCHMIDT,  RUDOLF.  The  Alternating  Current  Commutator  Motor.  The 
Leakage  of  Induction  Motors.  In  one  volume.  247  illustrations.  5^x8%, 
cloth Net,  $3.00 

GORE,  GEORGE.  The  Art  of  Electrolytic  Separation  of  Metils  (Theoretical  and 
Practical).  Illustrated.  8vo.,  cloth,  295  pp Net,  $3 . 50 

GROTH,  L.  A.  Welding  and  Cutting  Metals  by  Aid  of  Gases  or  Electricity.  124 
illustrations.  8vo.,  cloth,  280  pp.  (Van  Nostrand's  Westminster  Series.) 

Net,  $2.00 

HALLER,  G.  F.  and  CUNNINGHAM,  E.  T.  The  Tesla  High  Frequency  Coil;  its 
construction  and  uses.56  illustrations  5x7^,  cloth,  130  pp.. .......  .$1.25 

HASKINS,  C.  H.  The  Galvanometer  and  its  Uses.  A  Manual  for  Electricians 
and  Students.  Fifth  Edition,  revised.  Illus.  16mo.,  morocco,  75  pp. .  .$1 .50 

HAUSMANN,  E.  Telegraph  Engineering.  A  Manual  for  Practing  Telegraph 
Engineers  and  Engineering  Students.  192  illustrations.  5^x8^,  .cloth, 
416  pp , Net,  $2.00 

HAY,  ALFRED.  An  Introductory  Course  of  Continuous-Current  Engineering.  Illus- 
trated. 8vo.,  cloth,  327  pp Net,  $2.50 


6  LIST  OF  WORKS  ON  ELECTRICAL  SCIENCE. 

HAYES,  H.  V.  Public  Utilities,  Their  Cost  New  and  Depreciation.  5%xg^,  cloth, 
275  pp Net,  $2.00 

Public  Utilities;  Their  Fair  Present  Value  and  Return.     5%x8j^,  cloth,  200  pp. 

Net,  $2.50 

HEATHER,  H.  J.  Electrical  Engineering  for  Mechanical  and  Mining  Engineers. 
183  illustrations.  5%x8%,  cloth,  344  pp Net,  $3 . 50 

HEAVISIDE,  0.  Electromagnetic  Theory.  Two  Volumes  with  Many  Diagrams. 
Svo.,  cloth,  1006  pp.  Each  Vol Net,  $5.00 

Vol.  III.     529  pp Net,  $7.50 

HOBART,  H.  M.     Heavy  Electrical  Engineering.     Illustrated.     8vo.,   cloth,  338 
pp Net,  $4.50 

Electricity.      A  text-book    designed  in   particular  for  engineering  students. 

115  illustrations.     43  tables.     Svo.,  cloth,  266  pp Net,  $2.00 

Design   of    Static    Transformers.      100   Illustrations,   8vo.,    cloth,   190  pp. 

Net,  $2.00 

Electric  Trains.    88  illustrations.    8 vo.,  cloth,  220  pp Net.  $2 . 50 

Electric  Propulsion  of  Ships.  44  illustrations.  8vo.,  cloth,  167  pp. Net,  $2.00 

HOBBS,  W.  R.  P.  The  Arithmetic  of  Electrical  Measurements.  With  numerous 
examples,  fully  worked.  Sixteenth  Edition,  revised  and  edited  with  six 
additional  chapters  by  A.  R.  Palmer.  12mo.,  cloth,  126  pp 50  cents 

HOPKINS,  N.  M.  Experimental  Electrochemistry,  Theoretically  and  -Practically 
Treated.  Profusely  illustrated  with  130 -new  drawings,  diagrams,  and  photo- 
graphs, accompanied  by  a  Bibliography.  New  Edition.  In  Press. 

HOUSTOUN,  R.  A.  Studies  in  Light  Production.  22  illustrations.  5fx8f, 
cloth,  120  pp Net,  $2.00 

HUTCHINSON,  R.  W.,  Jr.  Long-Distance  Electric  Power  Transmission:  Being 
a  Treatise  on  the  Hydro-Electric  Generation  of  Energy;  Its  Transformation, 
Transmission,  and  Distribution.  Second  Edition.  Illustrated.  12mo., 
cloth,  350  pp Net,  $3.00 

and  THOMAS,  W.  D.  Electricity  in  Mining.  Being  a  theoretical  and  prac- 
tical treatise  on  the  construction,  operation,  and  maintenance  of  electrical 
mining  machinery.  Illustrated.  12mo.,  cloth In  Press 

INCANDESCENT  ELECTRIC  LIGHTING.  A  Practical  Description  of  the  Edison 
System,  by  H.  Latimer.  To  which  is  added:  The  Design  and  Operation  of 
Incandescent  Stations,  by  C.  J.  Field;  A  Description  of  the  Edison  Electro- 
lyte Meter,  by  A.  E.  Kennelly;  and  a  Paper  on  the  Maximum  Efficiency  of 
Incandescent  Lamps,  by  T.  W.  Ho  well.  Fifth  Edition.  Illustrated. 
16mo.,  cloth,  140  pp.  (No.  57  Van  Nostrand's  Science  Series.) 50  cents 

INDUCTION  COILS:    How  Made  and  How  Used.    Eleventh  Edition.     Illustrated 
16mo.,  cloth,  123  pp.     (No.  53  Van  Nostrand's  Science  Series.). .  .50  cents 


LIST  OF  WORKS  ON  ELECTRICAL  SCIENCE.  1 

JEHL,  FRANCIS.  The  Manufacture  of  Carbons  for  Electric  Lighting  and  Other 
Purposes.  Illustrated.  8vo.,  cloth,  232  pp Net,  $4 .00 

JOHNSON,  J.  H.  Arc  Lamps  and  Accessory  Apparatus.  20  Illustrations. 
16mo.,  cloth,  135  pp.  (Installation  Manuals  Series) Net,  .75 

JOHNSON,  T.  M.  Ship  Wiring  and  Fitting.  47  illustrations.  16mo.,  cloth,  92 
pp.  (Installation  Manuals  Series) Net,  75  cents. 

JONES,  HARRY  C.     The  Electrical  Nature  of  Matter  and  Radioactivity.      Third 
Edition,  completely  remsed.     5^x8%,  cloth.  218  pp $2 .00 

KAPP,  GISBERT.  Alternate-Current  Machinery.  Illustrated.  16mo.,  cloth,  190 
pp.  (No.  96  Van  Nostrand's  Science  Series.) 50  cents 

KENNEDY,  R.  Electrical  Installations  of  Electric  Light,  Power,  and  Traction 
Machinery.  Illustrated.  8vo.,  cloth,  5  vols.  The  Set,  $15. 00. Each,  $3.50 

KENNELLY,  A.  E.  Theoretical  Elements  of  Electro-Dynamic  Machinery.  Vol  I. 
Illustrated.  8vo.,  cloth,  90  pp $1 .50 

KERSHAW,  J.  B.  C.  The  Electric  Furnace  in  iron  and  Steel  Production.  Illus- 
trated. 8vo.,  cloth,  74  pp Net,  $1 .50 

Electrometallurgy.  Illustrated.  8 vo.,  cloth,  303  pp.  (Van  Nostrand's  West- 
minster Series.) Net,  $2 .00 

KINZBRUNNER,  C.  Continuous-Current  Armatures;  their  Winding  and  Con- 
struction. 79  Illustrations.  8vo.,  cloth,  80  pp Net,  $1 .50 

Alternate-Current  Windings;  their  Theory  and  Construction.  89  Illustrations. 

8vo.,  cloth,  80  pp Wet,  $1 .50 

The  Testing  of  Altercating  Current  Machines  in  Laboratories  and  Test  Rooms. 

A  practical  work  for  students  and  engineers.  Vol.  I.  General  Tests; 
Transformers,  Alternators.  141  illustrations.  5Jix8%,  cloth,  164  pp 

Net,  $2.00 

KOESTER,  F.  Hydroelectric  Developments  and  Engineering.  A  practical  and 
theoretical  treatise  on  the  development,  design,  construction,  equipment  and 
operation  of  hydroelectric  transmission  plants.  Second  Edition.  500  illus- 
trations. 4to.,  cloth,  475  pp .  ..Net,  $5.00 

• Steam-Electric  Power  Plants.    A  practical  treatise  on  the  design  of  central 

light  and  power  stations  and  their  economical  construction  and  operation. 
Fully  Illustrated.  Second  Edition  Ato. ,  cloth,  455  pp Net,  $5 . 00 

LANCASTER,  M.  Electric  Cooking,  Heating  and  Cleaning.  Edited  by  W.  E. 
Lancaster,  American  Edition  by  S.  L.  Coles,  305  illustrations,  6x8%,  cloth, 
340  pp Net,  $1.50 

LARNER,  E.  T.  The  Principles  of  Alternating  Currents  for  Students  of  Electrical 
Engineering.  Illustrated  with  Diagrams.  12mo.,  cloth,  144  pp. Net,  $1.25 

LEMSTROM,  S.  Electricity  in  Agriculture  and  Horticulture.  Illustrated.  8vo., 
cloth Net,  $1 .50 


8  LISI   OF   WORKS  ON  ELECTRICAL  SCIENCE. 

LIVERMORE,  V.  P.,  and  WILLIAMS,  J.  How  to  Become  a  Competent  Motorman: 
Being  a  practical  treatise  on  the  proper  method  of  operating  a  street-railway 
motor-car;  also  giving  details  how  to  overcome  certain  defects.  Second 
Edition.  Illustrated.  16mo.,  cloth,  247  pp Net,  $1 .00 

LIVINGSTONE,  R.  Mechanical  Design  and  Construction  of  Generators.  122 
illustrations.  5^x8%,  cloth.  22S  pp Net,  $3 .50 

Mechanical  Design  and  Construction  of  Commutators.  62  illustrations, 

5^x8^,  cloth,  93  pp Net,  2 .25 

LOCKWOOD,  T.  D.  Electricity,  Magnetism,  and  Electro-Telegraphy.  A  Prac- 
tical Guide  and  Handbook  of  General  Information  for  Electrical  Students, 
Operators,  and  Inspectors.  Fourth  Edition.  Illustrated.  8vo.,  cloth, 
374  pp $2.50 

LODGE,  OLIVER  J.  Signalling  Across  Space  Without  Wires :  Being  a  description 
of  the  work  of  Hertz  and  his  successors.  Fourth  Edition.  Illustrated.  8vo., 
cloth,  156  pp Net,  $2.00 

LORING,  A.  E.  A  Handbook  of  the  Electro-Magnetic  Telegraph.  Fourth  Edition, 
revised.  Illustrated.  16mo.,  cloth,  116  pp.  (No.  39  Van  Nostrand's 
Science  Series.) 50  cents 

LUCKIESH,  M.  Color  and  Its  Application.  126  illustrations,  4  color  plates. 
6x9,  cloth,  350  pp In  Press 

MALCOLM,  W.  H.     Theory  of  the  Submarine  Telegraph  Cable.  In  Press 

MANSFIELD,  A.  N.  Electromagnets:  Their  Design  and  Construction.  Second 
Edition.  Illustrated.  16mo.,  cloth,  155  pp.  (Van  Nostrand's  Science 
Series  No.  64) 50  cents 

Manufacture  of  Electric  Light  Carbons.     Illustrated.     5^x8,  cloth.  Net,  $1 .00 

MASSIE,  W.  W.,  and  UNDERHILL,  C.  R.  Wireless  Telegraphy  and  Telephony 
Popularly  Explained.  Illustrated.  12mo.,  cloth,  82  pp Net,  $1 .00 

MAURICE,  W.  Electrical  Blasting  Apparatus  and  Explosives,  with  special 
reference  to  colliery  practice.  Illustrated,  8vo.,  cloth,  167  pp.  .Net,  $3.50 

The  Shot  Firer's  Guide.     A  practical  manual  on  blasting  and  the    prevention 

of  blasting  and  accidents.  78  illustrations.   8vo.,  cloth,  212  pp. Net,  $1 .50 

MEISSNER,  B.  F.     Radio-Dynamics In  Press 

MONCKTON,  C.  C.  F.     Radio  Telegraphy.    173  illustrations.     8vo.,  cloth.  272  pp 

Net,  $2.00 

MONTGOMERY,  J.  W.      Electric  Wiring  Specifications.     4x6^,  cloth,  107  pp. 

Net,  $1.00 

MORECROFT,  J.  H.  and  HEHRE,  F.  W.  A  Short  Course  in  Testing  of  Electrical 
Machinery.  Third  Edition.  84  Illustrations.  8vo.,  cloth,  160  pp. Net,  SI. 50 

MORGAN,  ALFRED  P.  Wireless  Telegraph  Construction  for  Amateurs.  Third 
Edition,  167  illustrations.  12mo.,  cloth,  236  pp Net,  $1.50 


LIST  OF  WORKS  ON  ELECTRICAL  SCIENCE.  9 

NERZ,  F.  Searchlights,  Their  Theory,  Construction  and  Application.  Trans- 
lated by  C.  Rodgers.  47  Illustrations.  6x8,  cloth,  145  pp.. .  .Net,  $3.00 

NIPHER,  FRANCIS  E.  Theory  of  Magnetic  Measurements.  With  an  Appendix 
on  the  Method  of  Least  Squares.  Illustrated.  12mo.,  cloth,  94  pp.$l  .00 

OHM,    G.  S.      The  Galvanic   Circuit   Investigated  Mathematically.     Berlin,  1827 
Translated  by  William  Francis.     With  Preface  and  Notes  by  the   Editor, 
Thos.  D.  Lockwood.     Second  Edition.     Illustrated.     16mo.,  cloth,  269  pp. 
(No.  102  Van  Nosrrand's  Science  Series.) 50  cents 

OLSSON,  ANDREW.  Motor  Control  as  used  hi  Connection  with  Turret  Turning 
and  Gun  Elevating.  (The  Ward  Leonard  System.)  13  illustrations.  12mo., 
paper,  27  pp.  (U.  S.  Navy  Electrical  Series  No.  1.) Net,  50  cents 

OUDIN,  MAURICE  A.  Standard  Polyphase  Apparatus  and  Systems.  Illustrated 
with  many  Photo-reproductions,  Diagrams,  and  Tables.  Sixth  Edition, 
revised.  8vo.,  cloth.  369  pp Net,  $3 .  00 

PALAZ,  A.  Treatise  on  Industrial  Photometry.  Specially  applied  to  Electric 
Lighting.  Translated  from  the  French  by  G.  W.  Patterson,  Jr.,  Assistant 
Professor  of  Physics  in  the  University  of  Michigan,  and  M.  R.  Patterson, 
B.  A.  Second  Edition.  Fully  illustrated.  Svo.,  cloth,  324  pp $4.00 

PARR,  G.  D.  A.  Electricai  Engine 2. i-.g  Measuring  Instruments  for  Commercial 
and  Laboratory  Purposes.  With  370  Diagrams  and  Engravings.  8vo., 
cloth,  328  pp Net,  $3.50 

PARSHALL,  H.  F.  and  HOBART,  H.  M.  Armature  Windings  of  Electric  Machines. 
Third  Edition.  With  140  full-page  Plates,  65  Tables,  and  165  pages  of 
descriptive  letter-press.  4to.,  cloth,  300  pp $7.58 

Electric  Railway  Engineering.     With  437  Figures  and  Diagrams  and  many 

Tables.     4to.,  cloth,  475  pp Net,   $10 .00 

PATCHELL,  W.  H.  Application  of  Electric  Power  to  Mines  and  Heavy  Industries. 
91  Illustrations.  6£x9±,  cloth,  344  pp Net,  $4 .00 

PATTERSON,  G.  W.  L.  Wiring  Calculations  for  Light  and  Power  Installations. 
139  Illustrations.  5}x7*,  cloth,  203  pp Net,  $1 .60 

Electric  Mine  Signalling  and  Installations.  Illustrated.  5ix7*,  cloth,  110  pp. 

Net,  $2.00 

PERRINE,  F.  A.  C.  Conductors  for  Electrical  Distribution :  Their  Manufacture 
and  Materials,  the  Calculation  of  Circuits,  Pole-Line  Construction,  Under- 
ground Working,  and  other  Uses.  Second  Edition.  Illustrated.  8vo., 
cloth,  287  pp Net,  $3.50 

POPE,  F.  L.  Modern  Practice  of  the  Electric  Telegraph.  A  Handbook  for  Elec- 
tricians and  Operators.  Seventeenth  Edition.  Illustrated.  Svo.,  cloth, 
234  pp $1.50 

RAPHAEL,  F.  C.  Localization  of  Faults  in  Electric  Light  Mains.  Third  Edition, 
revised In  Press 

RAYMOND,  E.  B.  Alternating-Current  Engineering,  Practically  Treated.  Third 
Edition^  revised.  With  many  Figures  and  Diagrams.  8vo.,  cloth,  244  pp 

Net,  $2.50 


10  LIST  OF  WORKS  ON  ELECTRICAL  SCIENCE. 

REDFERN,  T.  B.  and  SAVIN,  J.  Bells,  Indicators,  Telephones,  Fire  and  Burglar 
Alarms.  85  illustrations.  4£x6f,  cloth,  123  pp.  (Installation  Manuals 
Series) 0.50 

RICHARDSON,  S.  S.  Magnetism  and  Electricity  and  the  Principles  of  Electrical 
Measurement.  254  illustrations.  12mo.,  cloth,  598  pp Net,  $2.00 

ROBERTS,  J.  Laboratory  Work  in  Electrical  Engineering— Preliminary  Grade.  A 
series  of  Laboratory  experiments  for  first  and  second-year  students  in 
electrical  Engineering  Illustrated  with  many  Diagrams.  8vo.,  cloth, 
218  pp Net,  $2.00 

ROLLINS,  W.  Notes  on  X-Light.  Printed  on  deckle  edge  Japan  paper.  400 
pp.  of  text ,  152  full-page  plates.  Svo.,  cloth $5.00 

RUHMER,  ERNST.  Wireless  Telephony  in  Theory  and  Practice.  Translated 
from  the  German  by  James  Erskine-Murray.  Illustrated.  8vo.,  Cloth, 
218  pp Net  $2  . 00 

RUSSELL,  A.  The  Theory  of  Electric  Cables  and  Networks.  71  illustrations. 
8vo.,  cloth,  275  pp Net,  $3.00 

SAYERS,  H.  M.     Brakes  for  Tramwav  Cars.     6x9,  cloth,    76  pp Net,  $1.25 

SEVER,  G.  F.  Electrical  Engineering  Experiments  and  Tests  on  Direct- Current 
Machinery.  Second  Edition,  enlarged.  With  Diagrams  and  Figures.  8vo., 
pamphlet,  75  pp Net,  $1.00 

SEVER,  G.  F.  and  TOWNSEND,  F.  Laboratory  and  Factory  Tests  in  Electrical 
Engineering.  Seeond  Edition.  Illustrated.  8vo.,  cloth,  269  pp.  .Net,  $2.50 

SEW  ALL,   C.    H.      Wireless  Telegraphy.     With  Diagrams  and  Figures.     Second 

Edition,  corrected.     Illustrated.     8vo.,  cloth,  229  pp Net,  $2.00 

Lessons  in  Telegraphy.     Illustrated.     12mo.,  cloth,  104  pp Net,  $1 .00 

SEW  ALL,  T.  The  Construction  of  Dynamos  (Alternating  and  Direct  Current).  A 
Textbook  for  students,  engineering  contractors,  and  Electricians-in-charge. 
Illustrated.  8vo.,  cloth,  316  pp $3.00 

SHELDON,  S.,  and  HAUSMANN,E.  Dynamo-Electric  Machinery:    Its  Construction, 

Design,  and  Operation. 
Vol.    I.:     Direct-Current  Machines.       Ninth    Edition,     completely    rewritten. 

Illustrated.     12mo.,  cloth,  281  pp Net,  $2.50 

Vol.  II.:     Alternating-Current  Machines:      Tenth   Edition,   rewritten.     12mo., 

cloth,  353  pp Net,  $2.50 

Electric  Traction  and  Transmission  Engineering.     127  illustration.      12mo., 

cloth.     317   pp Net,  $2 .50 

SLOANE,  T.  O'CONOR.     Standard  Electrical  Dictionary.     300  Illustrations.     12mo., 

cloth,  682  pp $3.00 

Elementary    Electrical    Calculations.        A    Manual    of  Simple    Engineering 

Mathematics,  covering  the  whole  field  of  Direct  Current  Calculations,  the 

basis  of  Alternating  Current  Mathematics,  Networks,  and  typical  cases  of 

Circuits,    with  Appendices  on  special  subject.      8vo.,  cloth.      Illustrated. 

304  pp Net,  $2 . 00 


LIST  OF  WORKS   ON  ELECTRICAL   SCIENCE.  11 

SMITH,  C.  F.     Practical  Alternating  Currents,  and  Alternating  Current  Testing. 

Third  Edition.  236  illustrations.  5^x894,  cloth.  476  pp Net,  $2.50 

SMITH,  C.  F.  Practical  Testing  of  Dynamos  and  Motors  Third  Edition.  108 

illustrations.  5^x8%,  cloth,  322  pp Net,  $2.00 

SNELL,  ALBION  T.  Electric  Motive  Power.  The  Transmission  and  Distribution 

of  Electric  Power  by  Continuous  and  Alternating  GUI  rents.     With  a  Section 

on    the   Applications   of   Electricity    to    Mining   Work.     Second   Edition. 

Illustrated.     8vo.,  cloth,  411  pp Net,  $4.00 

SODDY,  F.  Radio-Activity;  an  Elementary  Treatise  from  the  Standpoint  of  the 
Disintegration  Theory.  Fully  Illustrated.  8vo.,  cloth,  214  pp.  .Net,  $3.00 

SOLOMON,  MAURICE.  Electric  Lamps.  Illustrated.  8vo.,  cloth.  (Van  Nos- 
trand's  Westminster  Series.) Net,  $2 .00 

SWINBURNE,  JAS.,  and  WORDINGHAM,  C.  H.  The  Measurement  ofE  lectric 
Currents.  Electrical  Measuring  Instruments.  Meters  for  Electrical  Energy. 
Edited,  with  Preface,  by  T.  Commerford  Martin.  Folding  Plate  and  Numer- 
ous Illustrations.  16mo.,  cloth,  241  pp.  (No.  109  Van  Nostrand's  Science 
Series.) 50  cents 

SWOOPE,  C.  WALTON.  Lessons  in  Practical  Electricity:  Principle  Experi- 
ments, and  Arithmetical  Problems.  An  Elementary  Text-book.  Fifteenth 
Edition,  enlarged  with  a  chapter  on  electric  lighting.  404  illustrations. 
12mo.,  cloth,  462  pp Net,  $2.00 

THIESS,  J.  B.  and  JOY,  G.  A.  Toll  Telephone  Practice.  273  illustrations.  8vo., 
cloth,  433  pp Net,  $3.50 

THOM,  C.,  and  JONES,  W.  H.  Telegraphic  Connections,  embracing  recent  methods 
in  Quadruplex  Telegraphy.  20  Colored  Plates.  8vo.,  cloth,  59  pp.  .$1 .50 

THOMPSON,  S.  P.,  Prof.  Dynamo-Electric  Machinery.  With  an  Introduction 
and  Notes  by  Frank  L.  Pope  and  H.  R.  Butler.  Illustrated.  16mo., 
cloth,  214  pp.  (No.  66  Van  Nostrand's  Science  Series.) 50  cents 

Recent  Progress  in  Dynamo-Electric  Machines.  Being  a  Supplement  to 

"  Dynamo-Electric  Machinery."  Illustrated.  16mo.,  cloth,  113  pp.  (No. 
75  Van  Nostrand's  Science  Series.) 50  cents 

TOWNSEND,  FITZHUGH.  Alternating  Current  Engineering.  Illustrated.  8vo., 
paper,  32  pp Net,  75  cents 

UNDERBILL,  C.  R.      Solenoids,   Electromagnets  and  Electromagnetic  Windings. 

Second  Edition.  218  Illustrations.  12mo.,cloth,  345  pp Net,  $2.00 

URQUHART,  J.  W.  Electroplating.  Fith  Edition.  lUustrated.  12mo.,  cloth, 

230  pp $2.00 

Electrotyping.  Illustrated.  12mo.,  cloth,  228  pp $2.00 

VOSMAER,  A.  Ozone.  Its  Manufacture  and  Uses.  Illustrated.  In  Press 

WADE,  E.  J.  Secondary  Batteries:  Their  Theory,  Construction,  and  Use.  Second 
Edition,  corrected  265  illustrations.  8vo.,  cloth,  501  pp.  Net,  $4.00 


12  LIST  01  WORKS  ON  ELECTRICAL  SCIENCE."*  3 

WADSWORTH,  C.  Primary  Battery  Ignition.  A  simple  practical  pocket  guide 
on  the  construction,  operation,  maintenance,  and  testing  of  primary 
batteries  for  automobile,  motorboat,  and  stationary  engine  ignition  service. 
26  Illustrations.  5x7,  cloth,  79  pp Net,  0 .50 

WALKER,  FREDERICK.  Practical  Dynamo-Building  for  Amateurs.  How  to 
Wind  for  any  Output.  Third  Edition.  Illustrated.  16mo.,  cloth,  104  pp. 
(No.  68  Van  Nostrand's  Science  Series.) 50  cents. 

WALKER,  SIDNEY  F.     Electricity  in  Mining.     Illustrated.    8vo.,  cloth,  385  pp. 

$3.50 

WATT,  ALEXANDER.     Electroplating  and  Refining   of  Metals.      New  Edition, 

rewritten  by  Arnold  Philip.     Illustrated.     8vo.,  cloth,  704  pp. Net,  $4.50 

Electro-metallurgy    Fifteenth  Edilion.   Illustrated     12mo.,  cloth,  225  pp. $1.00 

WEBB,  H.  L.  A  Practical  Guide  to  the  Testing  of  Insulated  Wires  and  Cables. 
Fifth  Edition.  Illustrated.  12mo.,  cloth.,  118  pp $1 . 00 

WEYMOUTH,  F.  MARTEN.  Drum  Armatures  and  Commutators.  (Theory  and 
Practice.)  A  complete  treatise  on  the  theory  and  construction  of  drum- 
winding,  and  of  commutators  for  closed-coil  armatures,  together  with  a  full 
re'sume'  of  some  of  the  principal  points  involved  in  their  design,  and  an 
exposition  of  armature  reactions  and  sparking.  Illustrated.  8vo.,  cloth, 
295  pp Net,  $3 . 00 

WILKINSON,  H.  D.  Submarine  Cable-Laying,  Repairing,  and  Testing.  Second 
Edition,  completely  revised.  313  illustrations.  8vo.,  cloth,  580  pp.Net,  |6.00 

WILSON,  J.  F.  Essentials  of  Electrical  Engineering.  300  illustrations.  6x9, 
clot.n.  255  pp Net,  $2.50 

WRIGHT,  J.  Testing,  Fault  Localization  and  General  Hints  for  Linemen.  19  illus- 
trations. 16mo.,  cloth,  88  pp.  (Installation  Serious  Manuals.) .  Net,  50  cents 

YOUNG,  J.  ELTON.  Electiical  Testing  for  Telegraph  Engineers.  Illustrated. 
8vo.,  cloth,  264  pp Net,  $4.00 

ZEIDLER,  J.  and  LUSTGARTEN,  J.  Electric  Arc  Lamps  Their  principles,  con- 
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3Mar58MF 

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General  Library 
LD  21A-50m-8,'57                               University  of  Calif  orni  a 
(C8481slO)476B                                                Berkeley 

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UNIVERSITY  OF  CALIFORNIA  LIBRARY 


